First-Order ODEs

Abstract

Modelling of various problems in engineering, physics, chemistry, biology and economics allows formulating of differential equations, where a being searched function is expressed via its time changes (velocities). One of the simplest example is that given by a first-order ODE of the form

$$\displaystyle{ \frac{dy} {dt} = F(y), }$$

(2.1)

where F(t) is a known function, and we are looking for y(t). Here by t we denote time. In general, any given differential equation has infinitely many solutions. In order to choose from infinite solutions those corresponding to a studied real process, one should attach initial conditions of the form $y(t_{0}) = y_{0}$.

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Modelling of various problems in engineering, physics, chemistry, biology and economics allows formulating of differential equations, where a being searched function is expressed via its time changes (velocities). One of the simplest example is that given by a first-order ODE of the form

$$\displaystyle{ \frac{dy} {dt} = F(y), }$$

(2.1)

where F(t) is a known function, and we are looking for y(t). Here by t we denote time. In general, any given differential equation has infinitely many solutions. In order to choose from infinite solutions those corresponding to a studied real process, one should attach initial conditions of the form $y(t_{0}) = y_{0}$.

In general, there is no direct rule/recipe for construction of an ODE. Let y = y(t) be a dependence between t and y of the investigated process. We are going to monitor the difference $y(t + \Delta t) - y(t)$ caused by the disturbance $\Delta t$. Then, if we take

$$\displaystyle{\dot{y} \equiv \frac{dy} {dt} =\lim \limits _{\Delta t\rightarrow 0}\frac{y(t + \Delta t) - y(t)} {\Delta t},}$$

we obtain a differential equation, i.e. dependence of the process velocity in the point t governed by the function F(y).

There are also cases where a function y(t) appears under an integral and the obtained equation is called the integral equation, which in simple cases can be transformed to a differential equation.

2.1 General Introduction

A differential equation of the form

$$\displaystyle{ f\left (t,y, \frac{dy} {dt} \right ) = 0 }$$

(2.2)

is called the first-order ordinary differential equation, where t is the independent variable (here referred to time, but in general it can be taken as a space variable x), and y(t) is the unknown function to be determined. Observe that Eq. (2.2) is not solved with respect to its derivative dy∕dt. In many cases, however, one deals with the following differential equation

$$\displaystyle{ \frac{dy} {dt} = f(t,y), }$$

(2.3)

which is called the first-order ODE solved with respect to the derivative. Alternatively, one may deal often with the following form of first-order ODE

$$\displaystyle{ P(t,y)dt + Q(t,y)dy = 0, }$$

(2.4)

where P, Q are given functions.

We say that y = ϕ(t) is a solution to either (2.2) or (2.3) in an interval J, if

$$\displaystyle{ f\left (t,\phi (t), \frac{d\phi (t)} {dt} \right ) \equiv 0, }$$

(2.5)

or

$$\displaystyle{ \frac{d\phi (t)} {dt} = f(t,\phi (t)), }$$

(2.6)

for all t ∈ J.

One may also find a solution to Eq. (2.2) in the implicit form $\varphi (t,\phi (t))$, where ϕ(t) = y is a solution to Eq. (2.2). Solution in the form of $\varphi (t,\phi (t))$ is also referred to as the integral of Eq. (2.2).

A graph of solution y = ϕ(t) of Eq. (2.2) is called the integral curve of the studied differential equation. Projection of the solution graph onto the plane (t, y) is called the phase curve (or trajectory) of the investigated first-order ODE.

A problem related to finding a solution y = ϕ(t) satisfying the initial condition $y(t_{0}) = y_{0}$ is called the Cauchy problem.

If we take a point (t, y) for t ∈ J, then a tangent line passing through this point creates with the axis t an angle α, then $\text{tan}\alpha = f(t,y)$. A family of all tangent lines defines a direction field for the studied differential equation. If we draw a short line segment possessing the slope f(t, y) through each of representative collection of points (t, y), then all line segments constitute a slope field for the investigated ODE.

A curve constituting of points with the same slope field is called the isocline. In other words all integral curves passing through an isocline intersect the axis t with the same angle.

Example 2.1.

Prove that the function y = ϕ(t) given in the parametric form t = xe ^x, $y = e^{-x}$ satisfies the following differential equation

$$\displaystyle{(1 + ty)\frac{dy} {dt} + y^{2} = 0.}$$

We have

$$\displaystyle\begin{array}{rcl} (1 + ty)\frac{dy} {dt} + y^{2}& =\,& (1 + xe^{x}e^{-x})\frac{dy} {dx} \frac{dx} {dt} + e^{-2x} {}\\ & =\,& -(1 + x)e^{-x} \cdot \frac{1} {(1 + x)e^{x}} + e^{-2x} = 0, {}\\ \end{array}$$

which proves that ϕ(t) satisfies the studied equation. □ 

Example 2.2.

Construct a differential equation of a family of ellipses of the following canonical form

$$\displaystyle{\frac{x^{2}} {a^{2}} + \frac{y^{2}} {b^{2}} = 1,}$$

where 0 < b < a.

Acting by d∕dx on both sides of this algebraic equation yields

$$\displaystyle{ \frac{x} {a^{2}} + \frac{y\frac{dy} {dx}} {b^{2}} = 0.}$$

Solving both equations we get

$$\displaystyle{\sqrt{a^{2 } - x^{2}} \frac{dy} {dx} + \frac{b} {a}x = 0.}$$

 □ 

Example 2.3.

Construct a differential equation of the force lines of a dipole constituted by two electric charges $(+q,-q)$ located on the distance 2a, where the force lines satisfy the Coulomb algebraic equation of the form

$$\displaystyle{\frac{x + a} {r_{1}} -\frac{x - a} {r_{2}} = C,}$$

where: $r_{1}^{2} = (x + a)^{2} + y^{2},\quad r_{2}^{2} = (x - a)^{2} + y^{2}$.

A differentiation of the algebraic equation yields

$$\displaystyle{\frac{r_{1} - (x + a)\frac{dr_{1}} {dx} } {r_{1}^{2}} -\frac{r_{2} - (x - a)\frac{dr_{2}} {dx} } {r_{2}^{2}} = 0,}$$

and also

$$\displaystyle{\frac{dr_{1}} {dx} = \frac{x + a + y\frac{dy} {dx}} {r_{1}},\quad \frac{dr_{2}} {dx} = \frac{x - a + y\frac{dy} {dx}} {r_{2}}.}$$

Finally, after a few of transformations we get

$$\displaystyle{\left (\frac{x - a} {r_{2}^{3}} -\frac{x - a} {r_{1}^{3}} \right )\frac{dy} {dx} + \left ( \frac{1} {r_{2}^{3}} - \frac{1} {r_{1}^{3}}\right )y = 0.}$$

 □ 

Example 2.4.

How many solutions of the equation $(x - 1)\frac{dy} {dx} + y = 0$ defines the relation

$$\displaystyle{y(x - 1) = C,}$$

for each fixed $C \in \mathbb{R}$. Find the solutions associated with the initial conditions y(0) = 0, $y(0) = -1$, y(2) = 1. Define intervals of solution existence as well as the corresponding integral and phase curves.

First we verify that $\varphi (x) = \frac{C} {x-1}$ satisfies the given differential equation. We have $\varphi _{1}(x) = \frac{C} {x-1}$ with x ∈ (C, +∞) and $\varphi _{2}(x) = \frac{C} {x-1}$ with x ∈ (1, +∞).

The initial condition y(0) = 0 is satisfied by the solution y = 0. Its integral curve corresponds to the axis of abscissa, whereas its phase corresponds to a projection of the integral curve into the axis of ordinates, i.e. the point y = 0.

In the case of $y(0) = -1$ we find that C = 1. It means that the integral curve of this solution corresponds the hyperbola branch $y(x - 1) = 1$ for x ∈ (−∞, 1). The phase curve of this solution is the ray y < 0.

Finally, in the case y(2) = 1 we obtain C = 1. Integral curve of the solution $y = \frac{1} {1-x}$ is the hyperbola $y(x - 1) = 1$ branch, where x ∈ (1, +∞) phase curve is the ray y > 0. □ 

2.2 Separable Equation

The first-order differential equation of the form

$$\displaystyle{ \frac{dy} {dx} = f(x)g(y) }$$

(2.7)

is called a separable differential equation.

If g(C ₀) = 0 in the point y = C ₀, then the function y = C ₀ is the solution to Eq. (2.7). If g(y) ≠ 0, then the following relation is obtained

$$\displaystyle{ \int \frac{dy} {g(y)} -\int f(x)dx = C. }$$

(2.8)

Theorem 2.1.

Let the function f(x) and g(x) are continuously differentiable in the vicinity of points x = x ₀ , y = y ₀ respectively, where g(y ₀ ) ≠ 0. Therefore, there is a unique solution y = ϕ(x) of Eq. (2.7)with the attached initial condition $\phi (x_{0}) = y_{0}$ in the vicinity of the point x = x ₀ , satisfying the relationship

$$\displaystyle{\int \limits _{y_{0}}^{\phi (x)} \frac{dy} {g(y)} =\int \limits _{ x_{0}}^{x}f(x)dx.}$$

If we have the equation

$$\displaystyle{ \frac{dy} {dx} = f(ax + by + c), }$$

(2.9)

then introducing a new variable

$$\displaystyle{ z = ax + by + c, }$$

(2.10)

we get

$$\displaystyle{ \frac{dz} {dx} = bf(z) + a, }$$

(2.11)

i.e. the problem is reduced to Eq. (2.7).

One may use the following physical interpretation of the differential equation

$$\displaystyle{ \frac{dy} {dx} = f(y). }$$

(2.12)

Let us attach to each point y a vector of the length | f(y) | , which direction is defined by the axis 0y providing that f(y) > 0. Therefore, a set of all vectors defines a vector field. The points f(y) = 0 are called singular points of the vector field (or its equilibrium positions in the case when we deal with time). Having drawn the vector field of the given Eq. (2.12) one may draw schematically the integral curves.

Example 2.5.

Find a solution of the following differential equation

$$\displaystyle{x(1 + y^{2}) + y(1 + x^{2})\frac{dy} {dx} = 0.}$$

We transform the studied equation to the form

$$\displaystyle{\int \frac{xdx} {1 + x^{2}} +\int \frac{ydy} {1 + y^{2}} = 2\ln C}$$

and hence after integration we get

$$\displaystyle{\ln (1 + x^{2}) +\ln (1 + y^{2}) =\ln C,}$$

which means that

$$\displaystyle{(1 + x^{2})(1 + y^{2}) = C.}$$

 □ 

Example 2.6.

Solve the following ODE

$$\displaystyle{\frac{dy} {dx} + y = 2x + 1.}$$

In order to transform the given ODE into that of separable variables we introduce the following new variable

$$\displaystyle{y - 2x - 1 = z,}$$

and hence

$$\displaystyle{\frac{dz} {dx} + z + 2 = 0.}$$

Separating variables and integrating we get

$$\displaystyle{\int \frac{dz} {z + 2} +\int dx = 0,}$$

which means that

$$\displaystyle{\ln \vert z + 2\vert + x =\ln C_{0},\quad \vert z + 2\vert = C_{0}e^{-x},\quad C_{ 0} > 0.}$$

Observe that $z = -2$ satisfies the studied equation directly, and therefore, all its solutions are given by the following formula

$$\displaystyle{z = -2 + Ce^{-x},\quad C \in \mathbb{R},}$$

and finally we get

$$\displaystyle{y = 2x - 1 + Ce^{-x}.}$$

 □ 

In what follows we proceed with a few examples of real-world applications.

Example 2.7.

A particle of mass m is subjected to action of a constant force, and it moves with the constant acceleration a. The viscous damping of the surrounding medium is c. Find the particle velocity providing that v(0) = 0.

The second Newton law gives

$$\displaystyle{\frac{dv(t)} {dt} = \frac{ma - cv(t)} {m},}$$

or equivalently

$$\displaystyle{\frac{dv} {dt} = -\frac{c} {m}v + a.}$$

The trivial (time independent solution) is

$$\displaystyle{v(t) = \frac{m} {c} a,}$$

and hence all solutions are given by the formula

$$\displaystyle{v(t) = \frac{m} {c} a + Ce^{-\frac{c} {m}t}}$$

The initial condition allows to find $C = -\frac{m} {c} a$, and finally

$$\displaystyle{v(t) = \frac{m} {c} a\left (1 - e^{-\frac{c} {m}t}\right ),}$$

which means also that

$$\displaystyle{\lim \limits _{t\rightarrow \infty }v(t) = \frac{m} {c} a.}$$

 □ 

Example 2.8.

A meteorite of mass M starts to move from its rest position into the Earth centre linearly from the height h (Fig. 2.1). Determine the meteorite velocity, when it touches the Earth surface assuming the Earth radius R.

Fig. 2.1

Meteorite movement towards Earth centre

We denote by y = y(t) the meteorite distance from its movement beginning point y(0) = 0, and by h − y(t) we denote the meteorite distance from the Earth centre in time instant t. The meteorite is subjected to action of two forces: Ma and Mg. Owing to the Newton principle we have

$$\displaystyle{\frac{Ma} {R^{2}} = \frac{Mg} {(h - y)^{2}},}$$

and hence

$$\displaystyle{a = \frac{gR^{2}} {(h - y)^{2}}.}$$

Therefore,

$$\displaystyle{a = \frac{dv} {dt} = \frac{dv} {dy} \cdot \frac{dy} {dt} = \frac{dv} {dy}v,}$$

and the following governing ODE is obtained

$$\displaystyle{v\frac{dv} {dy} = \frac{gR^{2}} {(h - y)^{2}},}$$

or equivalently

$$\displaystyle{\frac{1} {2} \frac{d(v)^{2}} {dy} = \frac{gR^{2}} {(h - y)^{2}}.}$$

Integration of the obtained equation yields

$$\displaystyle{v^{2} = \frac{2gR^{2}} {h - y} + C.}$$

Taking into account y(0) = 0, we get $C = -\frac{2gR^{2}} {h}$, and finally

$$\displaystyle{v^{2} = \frac{2gR^{2}y} {h(h - y)}.}$$

On the Earth surface $y = h - R$, and we get

$$\displaystyle{v = \sqrt{2gR\left (1 - \frac{R} {h} \right )}}$$

Taking into account that $h\rightarrow \infty$, the last formula yields

$$\displaystyle{v = \sqrt{2gR}.}$$

 □ 

Example 2.9.

Two substances A and B undergo a chemical reaction yielding a substance C. We assume amount of the C substance by y(t) in the time instant t after the reaction, and we denote by α and β the amount of substance A and B, in the beginning of reaction, respectively. Find $\frac{dy} {dt}$ assuming that the reaction velocity is proportional to the product of reacting masses.

The governing equation is

$$\displaystyle{\begin{array}{*{10}c} \dfrac{dy} {dt} = p(\alpha - y)(\beta -y),&p > 0, \end{array} }$$

and p is the proportionality coefficient. Separation of the variables yields

$$\displaystyle{ \frac{dy} {y -\alpha }- \frac{dy} {y-\beta } = -p(\beta -\alpha )dt.}$$

After integration one gets

$$\displaystyle{\frac{y -\alpha } {y-\beta } = Ce^{-p(\beta -\alpha )t}.}$$

Taking into account the initial condition y(0) = 0 we obtain the constant $C = \alpha /\beta$, i.e.

$$\displaystyle{\frac{y -\alpha } {y-\beta } = \frac{\alpha } {\beta }e^{-p(\beta -\alpha )t},}$$

or equivalently

$$\displaystyle{y(t) = \alpha \beta \frac{1 - e^{-p(\beta -\alpha )t}} {\beta -\alpha e^{-p(\beta -\alpha )t}}.}$$

Observe that for β > α we have

$$\displaystyle{\lim \limits _{t\rightarrow \infty }y(t) = \alpha,}$$

whereas for β < α we obtain

$$\displaystyle{\lim \limits _{t\rightarrow \infty }y(t) =\lim \alpha \beta \frac{e^{p(\beta -\alpha )t} - 1} {\beta e^{p(\beta -\alpha )t} -\alpha } =\beta.}$$

In the case when α = β the governing equation is

$$\displaystyle{\frac{dy} {dt} = p(\alpha - y)^{2}.}$$

Separation of the variables of this equation and the integration allows to find the following dependence

$$\displaystyle{ \frac{1} {\alpha - y} = pt + C.}$$

Since y(0) = 0, therefore $C = 1/\alpha$. In this case the reaction B governed by the equation

$$\displaystyle{y(t) = \alpha \left (1 - \frac{1} {1 + \alpha pt}\right ),}$$

which for t → ∞ yield

$$\displaystyle{\lim \limits _{t\rightarrow \infty }y(t) = \alpha.}$$

2.3 Homogenous Equations

A function F(x, y) is called homogenous of order k, if for all σ > 0 the following property holds [208]

$$\displaystyle{ F(\sigma x,\sigma y) = \sigma ^{k}F(x,y) }$$

(2.13)

For instance the functions

$$\displaystyle{ \frac{x + y} {x - y},\quad \frac{x^{2} + xy} {y - x},\quad x^{2} + y^{2} + 2xy }$$

(2.14)

are homogenous of order k = 0, 1, 2, , respectively.

A differential equation

$$\displaystyle{ \frac{dy} {dx} = F(x,y) }$$

(2.15)

is called homogenous, if the function F(x, y) is of order zero.

Equation

$$\displaystyle{ F_{1}(x,y)dx + F_{2}(x,y)dy = 0 }$$

(2.16)

is called homogeneous, if the function F ₁, F ₂ are homogeneous of the same order.

In the case of a homogeneous equation the introduction of a new variable y = zx allows to get en equation with separable variables. One may use also polar coordinates $(\varrho,\varphi )$ and by substitution $x =\varrho \cos \varphi$, $y =\varrho \sin \varphi$ again an equation with separable variables is obtained.

It should be mentioned that the equation

$$\displaystyle{ \frac{dy} {dx} = F\left (\frac{a_{1}x + b_{1}y + c_{1}} {a_{2}x + b_{2}y + c_{2}}\right ) }$$

(2.17)

can also be transformed to a homogeneous equation through the following linear transformation

$$\displaystyle{ \begin{array}{cc} x = x_{0} + X,&y = y_{0} + Y, \end{array} }$$

(2.18)

where $(x_{0},y_{0})$ is the point of intersection of straight lines $a_{1}x + b_{1}y + c_{1} = 0$ and $a_{2}x + b_{2}y + c_{2} = 0$. If the lines do not intersect then $a_{1}/b_{1} = a_{2}/b_{2}$, and in this case Eq. (2.17) is transformed to that with separable variables using

$$\displaystyle{ a_{1}x + b_{2}y + c_{1} = X. }$$

(2.19)

The function G(x, y) is called quasi-homogenous of order k, if for certain α and β the following relation holds

$$\displaystyle{ G(\sigma ^{\alpha }x,\sigma ^{\beta }y) = \sigma ^{k}G(x,y), }$$

(2.20)

for all k > 0.

Exponents α, β are called weights. We say that x(y) has weight α(β), and for instance $7x^{2}y^{5}$ has the weight 2α + 5β.

Differential equation (2.15) is called quasi-homogeneous if the associated function F(x, y) is quasi-homogeneous with weights α and β of order β −α, i.e. $F(\sigma ^{\alpha }x,\sigma ^{\beta }y) = \sigma ^{\beta -\alpha }F(x,y)$.

A quasi-homogeneous differential equation can be reduced to a homogeneous one. However, in many practical cases one may use the direct variables change $y = zx^{\frac{\beta }{\alpha }}$ allowing to get an equation with separable variables.

Example 2.10.

Find a solution of the following ODE

$$\displaystyle{\frac{dy} {dx} = \frac{xy + y^{2}e^{-\frac{x} {y} }} {x^{2}}.}$$

We introduce the new variable y = zx, and obtain

$$\displaystyle{x\frac{dz} {dx} + z = z + z^{2}e^{-\frac{1} {z} },}$$

or equivalently

$$\displaystyle{\frac{e^{\frac{1} {z} }} {z^{2}} dz = \frac{dx} {x}.}$$

Integration of the last equation yields

$$\displaystyle{-e^{\frac{1} {z} } =\ln \vert x\vert - C,}$$

or equivalently

$$\displaystyle{e^{\frac{x} {y} } +\ln \vert x\vert = C.}$$

 □ 

Example 2.11.

Solve the following equation

$$\displaystyle{\frac{dy} {dx} = 2\left ( \frac{y + 1} {x + y - 2}\right )^{2}.}$$

We introduce the following variables

$$\displaystyle{\begin{array}{*{10}c} y + 1 = Y,&x - 3 = X,\end{array} }$$

and we get

$$\displaystyle{\frac{dY } {dX} = 2 \frac{Y ^{2}} {(X + Y )^{2}}.}$$

Now we introduce the following new variable

$$\displaystyle{Y = uX,}$$

and the following ODE is obtained

$$\displaystyle{X \frac{du} {dX} + u = \frac{2u^{2}} {(1 + u)^{2}},}$$

or equivalently

$$\displaystyle{\ln \vert u\vert + 2\text{arctan}u +\ln \vert X\vert =\ln C,}$$

which means that

$$\displaystyle{uX = C\exp (-2\text{arctan}u).}$$

In the original variable the solution is

$$\displaystyle{(y + 1)\exp \left (2\text{arctan}\frac{y + 1} {x - 3}\right ) = C.}$$

 □ 

Example 2.12.

Prove that integral curves of the equation

$$\displaystyle{[2x(x^{2}-axy +y^{2})-y^{2}\sqrt{x^{2 } + y^{2}}]dx+y[2(x^{2}-axy +y^{2})+x\sqrt{x^{2 } + y^{2}}]dy = 0}$$

are closed curves surrounding the coordinates origin for | a |  < 2.

Since the studied equation is homogenous, then we introduce polar coordinates to get

$$\displaystyle\begin{array}{rcl} & \varrho ^{3}[2(1 - a\sin \varphi \cos \varphi )\cos \varphi -\sin ^{2}\varphi ](\cos \varphi d\varrho -\varrho \sin \varphi d\varphi ) & {}\\ & +\varrho ^{3}[2\sin \varphi (1 - a\sin \varphi \cos \varphi ) +\cos ^{2}\varphi ](\sin \varphi d\varrho + \varrho \cos \varphi d\varphi ) = 0& {}\\ \end{array}$$

or equivalently

$$\displaystyle{2(1 - a\sin \varphi \cos \varphi )d\varrho + \varrho \sin \varphi d\varphi = 0.}$$

Separating the variables we obtain

$$\displaystyle{\frac{d\varrho } {\varrho } + \frac{\sin \varphi } {2 - a\sin 2\varphi }d\varphi = 0,}$$

and after integration we get

$$\displaystyle{\ln \varrho +\int \limits _{ 0}^{\varphi } \frac{\sin udu} {2 - a\sin 2u} =\ln \varrho _{0},\quad \varrho _{0} = \varrho (0),}$$

or equivalently

$$\displaystyle{\varrho = \varrho _{0}\exp \left (\int \limits _{0}^{\varphi } \frac{\sin u} {2 - a\sin 2u}du\right ).}$$

If we prove that the function $\int \limits _{0}^{\varphi } \frac{\sin udu} {2-a\sin 2u}$ is periodic regarding $\varphi$ with the period 2π, then $\varrho = \varrho (\varphi )$ for arbitrary $\varrho _{0} > 0$ is the 2π periodic function and its integral curve is closed. We have

$$\displaystyle\begin{array}{rcl} \int \limits _{0}^{\varphi +2\pi } \frac{\sin udu} {2 - a\sin 2u}& =& \int \limits _{0}^{2\pi } \frac{\sin udu} {2 - a\sin 2u} +\int \limits _{ 2\pi }^{\varphi +2\pi } \frac{\sin udu} {2 - a\sin 2u} {}\\ & =& \int \limits _{0}^{\pi } \frac{\sin udu} {2 - a\sin 2u} -\int \limits _{\pi }^{2\pi } \frac{\sin udu} {2 - a\sin 2u} +\int \limits _{ 0}^{\varphi } \frac{\sin (2\pi + u)du} {2 - a\sin 2(2\pi + u)} {}\\ & =& \int \limits _{0}^{\pi } \frac{\sin du} {2 - a\sin 2u} -\int \limits _{0}^{\pi } \frac{\sin udu} {2 - a\sin 2u} +\int \limits _{ 0}^{\varphi } \frac{\sin udu} {2 - a\sin 2u} =\int \limits _{ 0}^{\varphi } \frac{\sin udu} {2 - a\sin 2u} {}\\ \end{array}$$

which proves that $\varrho (\varphi ) = \varrho (\varphi + 2\pi )$. □ 

Example 2.13.

Solve the following differential equation

$$\displaystyle{\frac{dy} {dx} = \frac{4x^{6} - y^{4}} {2x^{4}y}.}$$

Let x(y) assign the weight α(β). Then

$$\displaystyle{F(x,y) = \frac{4\sigma ^{6\alpha }x^{6} -\sigma ^{4\beta }y^{4}} {2\sigma ^{4\alpha +\beta }x^{4}y} = \sigma ^{\beta -\alpha }\frac{4x^{6} - y^{4}} {2x^{4}y}.}$$

This equation is satisfied when

$$\displaystyle{6\alpha - 4\alpha -\beta = 4\beta - 4\alpha -\beta =\beta -\alpha }$$

which means that $\beta /\alpha = 3/2$. Therefore, it has been proved that the studied equation is homogeneous. In order to separate its values, the following variable is introduced $y = zx^{\frac{3} {2} }$. We obtain

$$\displaystyle{\frac{dz} {dx}x^{\frac{3} {2} } + \frac{3} {2}x^{\frac{1} {2} }z = \frac{4x^{6} - z^{4}x^{6}} {2x^{4}zx^{\frac{3} {2} }},}$$

and multiplying both sides by $x^{-\frac{1} {2} }$ we get

$$\displaystyle{x\frac{dz} {dx} + \frac{3} {2}z = \frac{4 - z^{4}} {2z} }$$

or equivalently

$$\displaystyle{ \frac{2zdz} {(z^{2} + 4)(z^{2} - 1)} + \frac{dx} {x} = 0,\quad z\neq \pm 1.}$$

Direct integration yields

$$\displaystyle{ \ln \frac{\vert z^{2} - 1\vert } {z^{2} + 4} + 5\ln \vert x\vert =\ln C. }$$

(*)

In order to verify the obtained result we use the following differentiation formulas:

$$\displaystyle{(\ln \vert y\vert )^{{\prime}} = \frac{y^{{\prime}}} {y},\quad y = \frac{u} {v},\quad u = z^{2} - 1,\quad v = z^{2} + 4.}$$

Since

$$\displaystyle{y^{{\prime}} = \left (\frac{u} {v}\right )^{{\prime}} = \frac{vu^{{\prime}}- uv^{{\prime}}} {v} = \frac{2z(z^{2} + 4) - 2z(z^{2} - 1)} {(z^{2} + 4)} = \frac{10z} {(z^{2} + 4)^{2}},}$$

hence

$$\displaystyle{\frac{y^{{\prime}}} {y} = \frac{10z} {(z^{2} + 4)(z^{2} - 1)}.}$$

Full differentiation of (*) yields

$$\displaystyle{ \frac{10zdz} {(z^{2} + 4)(z^{2} - 1)} + \frac{5} {x}dx = 0.}$$

Formula (*) yields

$$\displaystyle{\frac{z^{2} - 1} {z^{2} + 4}x^{5} = C.}$$

Since $z^{2} = y^{2}/x^{3}$, therefore

$$\displaystyle{ \frac{y^{2} - x^{3}} {y^{2} + 4x^{3}}x^{5} = C.}$$

2.4 Linear Equations

Linear first-order ODE has the following form

$$\displaystyle{ \frac{dy} {dx} + a(x)y = f(x). }$$

(2.21)

There exist three different methods yielding a solution of Eq. (2.21)

• (i) The Lagrange method. This method is based on a constant variation. We consider first a homogeneous equation associated with (2.21) of the form

  $$\displaystyle{ \frac{dy} {dx} + a(x)y = 0, }$$

  (2.22)

  and its solution is

  $$\displaystyle{ y = C\exp \left [-\int a(x)dx\right ]. }$$

  (2.23)

  We are looking for solution to Eq. (2.21) by variation of the constant C = C(x), namely

  $$\displaystyle{ y = C(x)\exp \left [-\int a(x)dx\right ].L }$$

  (2.24)

  Substituting (2.24) to (2.21) we obtain

  $$\displaystyle{ \frac{dC(x)} {dx} = f(x)\exp \left [\int a(x)dx\right ], }$$

  (2.25)

  and hence

  $$\displaystyle{ C(x) = C +\int \left [f(x)\exp \left [\int a(x)dx\right ]\right ]dx, }$$

  (2.26)

  where C is the arbitrary constant.

  Finally, substitution of (2.26) into (2.24) yields

  $$\displaystyle{ y =\exp \left [-\int a(x)dx\right ]\left \{C +\int \left [f(x)\exp \left (\int a(x)dx\right )\right ]\right \}. }$$

  (2.27)

  Any solution passing through the point $(x_{0},y_{0})$ can be written in the following form

  $$\displaystyle{ y =\exp \left [-\int \limits _{x_{0}}^{x}a(z)dz\right ]\left \{y_{ 0} +\int \limits _{ x_{0}}^{x}f(u)\left [\exp \left (\int \limits _{ x_{0}}^{u}a(x)dx\right )du\right ]\right \}. }$$

  (2.28)

• (ii) The Bernoulli method. We are looking for a solution of (2.21) in the following form

  $$\displaystyle{ y = u(x)v(x). }$$

  (2.29)

  Substitution of (2.29) to (2.21) gives

  $$\displaystyle{ \frac{du} {dx}v + u\frac{dv} {dx} + a(x)uv = f(x). }$$

  (2.30)

  If we take u(x) as the solution of equation

  $$\displaystyle{ \frac{du} {dx} + a(x)u = 0, }$$

  (2.31)

  then

  $$\displaystyle{ u(x) =\exp \left [-\int a(x)dx\right ]. }$$

  (2.32)

  Substituting (2.32) into (2.30) gives

  $$\displaystyle{ \exp \left [-\int a(x)dx\right ]\frac{dv} {dx} = f(x), }$$

  (2.33)

  and therefore

  $$\displaystyle{ v(x) = C +\int f(x)\exp \left [\int a(x)dx\right ]dx, }$$

  (2.34)

  where C is a constant.

• (iii) The method of an integrating multiplier. We multiply both parts of Eq. (2.21) by $\exp \left (\int a(x)dx\right )$, and we get

  $$\displaystyle{ \frac{d} {dx}\left [y\exp \left (\int a(x)dx\right )\right ] = f(x)\exp \left (\int a(x)dx\right ) }$$

  (2.35)

  or equivalently

  $$\displaystyle{ y =\exp \left (-\int a(x)dx\right )\left [C +\int f(x)\exp \left (\int a(x)dx\right )dx\right ] }$$

  (2.36)

  Equation of the form

  $$\displaystyle{ A(y) + [B(y)x - C(y)]\frac{dy} {dx} = 0 }$$

  (2.37)

  can be transformed to the form (2.21). We multiply both sides by $\frac{1} {A} \frac{dx} {dy}$ and we get

  $$\displaystyle{ \frac{dx} {dy} + \alpha (y)x =\beta (y) }$$

  (2.38)

  where

  $$\displaystyle{ \alpha (y) = \frac{B} {A},\quad \beta (y) = \frac{C} {A}. }$$

  (2.39)

  It should be emphasized that equations of the form

  $$\displaystyle{ F^{{\prime}}(y)\frac{dy} {dx} + F(y)a(x) = b(x), }$$

  (2.40)

  where^′ denotes $\frac{d} {dy}$ can be transformed to the linear equation by introduction of the relation u = f(y).

A particular role in theory of first-order differential equations play the Bernoulli and Riccati equations. The equation

$$\displaystyle{ \begin{array}{cc} \dfrac{dy} {dx} + a(x)y = b(x)y^{n},&n\neq 0,1\end{array} }$$

(2.41)

is called the Bernoulli equation. It is transformed to the following form

$$\displaystyle{ \begin{array}{cc} y^{-n}\dfrac{dy} {dx} + a(x)y^{1-n} = b(x),&y\neq 0,\end{array} }$$

(2.42)

and it is reduced to a linear equation via the variable change $u = y^{1-n}$. This approach will be illustrated through examples. One may also apply here the Bernoulli method.

The equation

$$\displaystyle{ \frac{dy} {dx} + a(x)y + b(x)y^{2} = C(x) }$$

(2.43)

is called a Riccati equation. In general it cannot be solved in quadratures. However, if one of its particular solutions is known, say y ₁(x) then the transformation $y = y_{1} + u$ allows reduction of the problem to that of finding solution to the Bernoulli equation.

Example 2.14.

A current in the electrical network with the resistance R, induction L and excitation voltage u(t) = u ₀sinω t is governed by the following equation

$$\displaystyle{L\frac{di} {dt} + Ri = u_{0}\sin \omega t,\quad i(0) = 0.}$$

Find i = i(t).

We have

$$\displaystyle{\frac{di} {dt} + \alpha i =\beta \sin \omega t,}$$

where $\alpha = \frac{R} {L}$, $\beta = \frac{u_{0}} {L}$. We apply here the Bernoulli method, i.e. we assume

$$\displaystyle{i(t) = u(t)v(t).}$$

Substitution of i(t) into the governing equation yields

$$\displaystyle{ \frac{du} {dt} v + u\frac{dv} {dt} + \alpha uv =\beta \sin \omega t. }$$

(*)

We consider a solution of the homogeneous equation

$$\displaystyle{\frac{du} {dt} + \alpha u = 0}$$

of the form

$$\displaystyle{u =\exp (-\alpha t).}$$

We substitute it to (*) and we obtain $\frac{dv} {dt} =\beta e^{\alpha t}\sin \omega t$, what means that

$$\displaystyle{v(t) =\beta \left [\int e^{\alpha t}\sin \omega tdt + C\right ].}$$

We successively compute

$$\displaystyle\begin{array}{rcl} V (t)& =\,& \int e^{\alpha t}\sin \omega tdt = -e^{\alpha t}\frac{1} {\omega } \cos \omega t + \frac{\alpha } {\omega }\int e^{\alpha t}\cos \omega tdt {}\\ & =\,& -\frac{1} {\omega } e^{\alpha t}\cos \omega t + \frac{\alpha } {\omega ^{2}}e^{\alpha t}\sin \omega t -\frac{\alpha ^{2}} {\omega ^{2}}\int e^{\alpha t}\sin \omega tdt {}\\ & =\,& -\frac{1} {\omega } e^{\alpha t}\cos \omega t + \frac{\alpha } {\omega ^{2}}e^{\alpha t}\sin \omega t -\frac{\alpha ^{2}} {\omega ^{2}}V (t), {}\\ \end{array}$$

and therefore

$$\displaystyle{V (t) = \frac{e^{\alpha t}(-\omega \cos \omega t + \alpha \sin \omega t)} {\omega ^{2} + \alpha ^{2}}.}$$

Finally, we find

$$\displaystyle{v(t) =\beta \left [\frac{e^{\alpha t}(-\omega \cos \omega t + \alpha \sin \omega t)} {\omega ^{2} + \alpha ^{2}} + C\right ],}$$

and

$$\displaystyle{i(t) = u(t)v(t) =\beta \left (\frac{-\omega \cos \omega t + \alpha \sin \omega t} {\omega ^{2} + \alpha ^{2}} + Ce^{-\alpha t}\right ).}$$

Since i(0) = 0, $C = \frac{\omega } {\omega ^{2 } +\alpha ^{2}}$, and therefore

$$\displaystyle{i(t) = \frac{\beta } {\omega ^{2} + \alpha ^{2}}(-\omega \cos \omega t + \alpha \sin \omega t + \omega e^{-\alpha t}).}$$

Observe that

$$\displaystyle{\lim \limits _{t\rightarrow \infty }i(t) = \frac{u_{0}} {L(\omega ^{2} + \alpha ^{2})}(-\omega \cos \omega t + \alpha \sin \omega t) = \frac{u_{0}} {\sqrt{(L\omega )^{2 } + R^{2}}}\sin (\omega t -\varphi ),}$$

where $\text{tan}\varphi = \frac{\omega }{\alpha }$ denotes the initial current phase. □ 

Example 2.15.

Show that equation

$$\displaystyle{\frac{dy} {dx} + \alpha y = f(x),\quad \alpha > 0,}$$

possesses only one bounded solution assuming that f(x) is bounded for all $x \in \mathbb{R}$. Find this solution, and show that if $f(x + x_{0}) = f(x)$, then $y(x) = y(x + x_{0})$, where x ₀ is a period.

First we find a solution to the homogeneous equation

$$\displaystyle{\frac{dy} {dx} + \alpha y = 0.}$$

After variables separation we get

$$\displaystyle{\frac{dy} {y} = -\alpha dx,}$$

and hence

$$\displaystyle{\ln \vert y\vert + \alpha x =\ln C,}$$

which means that

$$\displaystyle{y = Ce^{-\alpha x}}$$

assuming that y ≠ 0.

We apply here the Lagrange’s method. Namely, we have

$$\displaystyle{y(x) = C(x)e^{-\alpha x},}$$

and substitution of y(x) into the governing equation gives

$$\displaystyle{\frac{dC} {dx} = e^{\alpha x}f(x).}$$

It means that

$$\displaystyle{C(x) = C(x_{0}) +\int \limits _{ x_{0}}^{x}e^{\alpha z}f(z)dz.}$$

The sought solution has the following form

$$\displaystyle{ y(x) = C(x_{0})e^{-\alpha x} +\int \limits _{ x_{0}}^{x}e^{-\alpha (x-z)}f(z)dz. }$$

(*)

Assuming $y(x_{0}) = y_{0}$ we obtain

$$\displaystyle{y_{0} = C(x_{0})e^{-\alpha x_{0} },}$$

or equivalently

$$\displaystyle{C(x_{0}) = y_{0}e^{\alpha x_{0} }.}$$

Therefore, solution (*) takes the following form

$$\displaystyle{y(x) = e^{-\alpha (x-x_{0})}y_{ 0} +\int \limits _{ x_{0}}^{x}e^{-\alpha (x-z)}f(z)dz.}$$

We multiply both sides of the last equation by $e^{\alpha (x-x_{0})}$ to get

$$\displaystyle{e^{\alpha (x-x_{0})}y(x) = y_{ 0} +\int \limits _{ x_{0}}^{x}e^{\alpha (z-x_{0})}f(z)dz.}$$

We consider the case $x\rightarrow -\infty$ (the case of $x\rightarrow + \infty$ can be studied in the similar way). We have

$$\displaystyle{\lim \limits _{x\rightarrow -\infty }e^{\alpha (x-x_{0})}y(x) = y_{ 0} +\int \limits _{ x_{0}}^{-\infty }e^{\alpha (z-x_{0})}f(z)dz,}$$

and hence

$$\displaystyle{y_{0} =\int \limits _{ -\infty }^{x_{0} }e^{\alpha (z-x_{0})}f(z)dz,}$$

because for a bounded solution $\lim \limits _{x\rightarrow -\infty }e^{\alpha (x-x_{0})}y(x) = 0.$ It means that

$$\displaystyle{ \lim \limits _{x\rightarrow -\infty }y(x) = Y (x) =\int \limits _{ -\infty }^{x}\exp (-\alpha (x - z))f(z)dz }$$

(**)

is bounded, assuming that f(z) is bounded.

In what follows we show that Y (x) is the only bounded solution of the studied equation. Let us assume that there exists one more bounded solution denoted by Y _∗(x). It means that the difference

$$\displaystyle{\Delta Y = Y (x) - Y _{{\ast}}(x)}$$

is bounded. We also have

$$\displaystyle\begin{array}{rcl} \frac{dY } {dx} + \alpha Y & =\,& f(x), {}\\ \frac{dY _{{\ast}}} {dx} + \alpha Y _{{\ast}}& =\,& f(x), {}\\ \end{array}$$

which means that

$$\displaystyle{\frac{d(\Delta Y )} {\Delta Y } + \alpha (\Delta Y ) = 0,}$$

and hence

$$\displaystyle{\Delta Y (x) = Ce^{\alpha x}.}$$

Owing to our introduced assumption $\Delta Y (x)$ is bounded for all $x \in \mathbb{R}$, which means that Ce ^{α x} must be bounded. This is true only if C = 0, which yields Y (x) = Y _∗(x).

Let us now show that if $f(x + x_{0}) = f(x)$ than $Y (x) = Y (x + x_{0})$. It follows from (**) that

$$\displaystyle\begin{array}{rcl} Y (x + x_{0})& =\,& \int \limits _{-\infty }^{x+x_{0} }e^{-\alpha (x+x_{0}-z)}f(z)dz =\int \limits _{ -\infty }^{x}e^{-\alpha (x-\tau )}f(\tau +x_{ 0})d\tau {}\\ & =\,& \int \limits _{-\infty }^{x}e^{-\alpha (x-\tau )}f(\tau )d\tau = Y (x). {}\\ \end{array}$$

 □ 

Example 2.16.

Solve the following Bernoulli equation

$$\displaystyle{x\frac{dy} {dx} + y = y^{2}\ln x.}$$

We use the Bernoulli method, and we look for a solution of the form

$$\displaystyle{y = u(x)v(x).}$$

Substitution of y(x) into the studied equation yields

$$\displaystyle{ xu\frac{dv} {dx} + xv\frac{du} {dx} + uv = u^{2}v^{2}\ln x }$$

(*)

We take

$$\displaystyle{x\frac{du} {dx} + u = 0,}$$

and hence

$$\displaystyle{u = \frac{1} {x}.}$$

Substituting this result into (*) we get

$$\displaystyle{x^{2} \frac{dv} {dx} = v^{2}\ln x,}$$

and hence

$$\displaystyle{ \frac{1} {v^{2}}dv = \frac{1} {x^{2}}\ln xdx.}$$

Integration of the last obtained equation gives

$$\displaystyle{-\frac{1} {v} =\int \frac{\ln x} {x^{2}}dx = -\frac{\ln x} {x} -\frac{1} {x} - C,}$$

which means that

$$\displaystyle{v(x) = \frac{x} {1 + Cx +\ln x}\,,}$$

and finally

$$\displaystyle{y(x) = u(x)v(x) = \frac{1} {1 + Cx +\ln x}.}$$

 □ 

Example 2.17.

Solve the following Bernoulli equation

$$\displaystyle{(1 + x^{2})\frac{dy} {dx} - 2xy = 4\sqrt{y(1 + x^{2 } )}\text{arctan}x.}$$

Assuming

$$\displaystyle{y(x) = u(x)v(x)}$$

we get

$$\displaystyle{(1 + x^{2})\left (\frac{du} {dx}v + u\frac{dv} {dx}\right ) - 2xuv = 4\sqrt{uv(1 + x^{2 } )}\text{arctan}x}$$

or equivalently

$$\displaystyle{(1 + x^{2})\frac{du} {dx}v + (1 + x^{2})\left (\frac{dv} {dx} - \frac{2x} {1 + x^{2}}v\right )u = 4\sqrt{uv(1 + x^{2 } )}\text{arctan}x.}$$

We take an arbitrary solution to the equation

$$\displaystyle{\frac{dv} {dx} - \frac{2x} {1 + x^{2}}v = 0,}$$

i.e. for example the following one

$$\displaystyle{v(x) = 1 + x^{2}.}$$

Therefore, we get

$$\displaystyle{(1 + x^{2})^{2}\frac{du} {dx} = 4(1 + x^{2})\sqrt{u}\text{arctan}x.}$$

One of the solution is u = 0, and the other solutions are found through the successive transformations

$$\displaystyle\begin{array}{rcl} \frac{du} {dx}& =\,& \frac{4\text{arctan}x} {1 + x^{2}} \sqrt{u}, {}\\ \frac{du} {2\sqrt{u}}& =\,& \frac{2\text{arctan}x} {1 + x^{2}} dx, {}\\ \sqrt{u}& =\,& \text{arctan}^{2}x + C. {}\\ \end{array}$$

Finally, the solutions are

$$\displaystyle\begin{array}{rcl} y& =\,& 0, {}\\ y& =\,& (1 + x^{2})(\text{arctan}^{2}x + C)^{2}. {}\\ \end{array}$$

 □ 

Example 2.18.

Solve the following Riccati equation

$$\displaystyle{\frac{dy} {dx} + y^{2} = \frac{2} {x^{2}}.}$$

Let us look for a particular solution of the form

$$\displaystyle{y_{1} = \frac{A} {x}.}$$

Substituting y ₁ into the studied equation yields

$$\displaystyle{-\frac{A} {x^{2}} + \frac{A^{2}} {x^{2}} = \frac{2} {x^{2}}.}$$

The second-order algebraic equation yields two roots $A_{1} = -1$, A ₂ = 2. Let us introduce a new variable z of the form

$$\displaystyle{y = z -\frac{1} {x},}$$

and therefore

$$\displaystyle{\frac{dz} {dx} + \frac{1} {x^{2}} + z^{2} -\frac{2z} {x} + \frac{1} {x^{2}} = \frac{2} {x^{2}},}$$

or equivalently

$$\displaystyle{\frac{dz} {dx} -\frac{2} {x}z = -z^{2}.}$$

We multiply both sides of the obtained equation by x ² to get

$$\displaystyle{x \frac{d} {dx}(zx) = 3zx - (zx)^{2}.}$$

We take

$$\displaystyle{zx = u,}$$

and integrate the following equation

$$\displaystyle{x\frac{du} {dx} = u(3 - u).}$$

Separation of the variables yields

$$\displaystyle{ \frac{du} {u(3 - u)} = \frac{dx} {x}.}$$

Since

$$\displaystyle{ \frac{1} {u(3 - u)} = \frac{1} {3u} + \frac{1} {3(3 - u)},}$$

therefore

$$\displaystyle{\frac{1} {3}\int \frac{du} {u} + \frac{1} {3}\int \frac{du} {3 - u} =\int \frac{dx} {x},}$$

and consequently

$$\displaystyle{\frac{1} {3}\left [\ln \vert u\vert -\ln \vert 3 - u\vert \right ] =\ln \vert x\vert +\ln C_{1},\qquad C_{1} > 0.}$$

Finally, we find

$$\displaystyle{\ln \left \vert \frac{u} {3 - u}\right \vert = 3\ln \vert C_{1}x\vert,}$$

or

$$\displaystyle{\ln \left \vert \frac{u} {3 - u}\right \vert =\ln \vert (C_{1}x)^{3}\vert.}$$

We consider two cases:

• (i)
  $$\displaystyle{ \frac{u} {3 - u} \geq 0.}$$

  In this case we have

  $$\displaystyle{ \frac{u} {3 - u} = Cx^{3},}$$

  which means that

  $$\displaystyle{ \frac{zx} {3 - zx} = Cx^{3},}$$

  and hence

  $$\displaystyle{z = \frac{3Cx^{2}} {Cx^{3} + 1}.}$$

  We finally get

  $$\displaystyle{y = z -\frac{1} {x} = \frac{2} {x}}$$

  and

  $$\displaystyle{y = \frac{3Cx^{2}} {Cx^{3} + 1} -\frac{1} {x} = \frac{2Cx^{3} - 1} {x(1 + Cx^{3})}.}$$
• (ii)
  $$\displaystyle{ \frac{u} {3 - u} < 0.}$$

  In this case we have

  $$\displaystyle{ \frac{u} {u - 3} = Cx^{3},}$$

  which means that

  $$\displaystyle{zx = C(zx - 3)x^{3},}$$

  and hence

  $$\displaystyle{z = \frac{3Cx^{2}} {Cx^{3} - 1}.}$$

  We finally obtain

  $$\displaystyle{y = \frac{2} {x}}$$

  and

  $$\displaystyle{y = \frac{2Cx^{3} + 1} {x(Cx^{3} - 1)}}$$

 □ 

2.5 Exact Differential Equations

The differential equation

$$\displaystyle{ M(x,y)dx + N(x,y)dy = 0 }$$

(2.44)

is called an exact differential equation if its left-hand side is the full differential of a certain function V (x, y) such that

$$\displaystyle{ dV (x,y) \equiv \frac{\partial V } {\partial x} dx + \frac{\partial V } {\partial y} dy = M(x,y)dx + N(x,y)dy = 0. }$$

(2.45)

A necessary condition that Eq. (2.44) is exact one follows

$$\displaystyle{ \frac{\partial M(x,y)} {\partial y} = \frac{\partial N(x,y)} {\partial x}. }$$

(2.46)

If V (x, y) is known than all solutions of (2.44) satisfy the condition

$$\displaystyle{ V (x,y) = C, }$$

(2.47)

where C is an arbitrary constant.

We show how we can find the function V (x, y). Since

$$\displaystyle{ \frac{\partial V } {\partial x} = M(x,y),\quad \frac{\partial V } {\partial y} = N(x,y), }$$

(2.48)

then

$$\displaystyle{ V (x,y) =\int M(x,y)dx =\uppsi (x,y) +\uppsi (y). }$$

(2.49)

We differentiate (2.49) to get

$$\displaystyle{ \frac{\partial \uppsi (x,y)} {\partial y} + \frac{\partial \uppsi (y)} {\partial y} = N(x,y). }$$

(2.50)

In some cases the general form given by (2.44) can be transformed to an exact differential equation by introduction of a so-called integrating multiplier m(x, y) [208]. In Eq. (2.46) we introduce m(x, y), and we obtain the following exact differential equation

$$\displaystyle{ \frac{\partial } {\partial y}(mM) = \frac{\partial } {\partial x}(mN), }$$

(2.51)

which means that m should satisfy the following equation

$$\displaystyle{ m\left (\frac{\partial M} {\partial y} -\frac{\partial N} {\partial x} \right ) = N \frac{\partial m} {\partial x} - M \frac{\partial m} {\partial y}. }$$

(2.52)

The obtained general form (2.52) can be simplified in the following cases

• (i) If m(x, y) = m(x) then

  $$\displaystyle{ \frac{1} {m} \frac{dm} {dx} = \frac{\frac{\partial M} {\partial y} -\frac{\partial N} {\partial x} } {N}. }$$

  (2.53)

• (ii) If m(x, y) = m(y) then

  $$\displaystyle{ -\frac{1} {m} \frac{dm} {dx} = \frac{\frac{\partial M} {\partial y} -\frac{\partial N} {\partial x} } {M}. }$$

  (2.54)

• (iii) If m(x, y) = m(r(x, y)), where r(x, y) is a known function then

  $$\displaystyle{ \frac{1} {m} \frac{dm} {dr} = \frac{\frac{\partial M} {\partial y} -\frac{\partial N} {\partial x} } {N \frac{\partial r} {\partial x} - M \frac{\partial r} {\partial y}}. }$$

  (2.55)

Example 2.19.

Solve the differential equation

$$\displaystyle{(2xy + 3y^{2})dx + (x^{2} + 6xy - 3y^{2})dy = 0.}$$

We have

$$\displaystyle{\begin{array}{*{10}c} M(x,y) = 2xy + 3y^{2},&N(x,y) = x^{2} + 6xy - 3y^{2}, \end{array} }$$

and hence

$$\displaystyle{\frac{\partial M} {\partial y} = \frac{\partial N} {\partial x} = 2x + 6y.}$$

It means that the left-hand side of the differential equation is a full differential of a certain function V (x, y).

We have

$$\displaystyle{\frac{\partial V } {\partial x} = 2xy + 3y^{2},\quad \frac{\partial V } {\partial y} = x^{2} + 6xy - 3y^{2}.}$$

First equation of the above yields

$$\displaystyle{V (x,y) = x^{2}y + 3xy^{2} +\uppsi (y).}$$

We differentiate the last equation with respect to y and thus

$$\displaystyle{\frac{\partial V } {\partial y} = x^{2} + 6xy + \frac{\partial \uppsi (y)} {\partial y} = x^{2} + 6xy - 3y^{2}.}$$

It means that

$$\displaystyle{\uppsi (y) = -y^{3} + C.}$$

Hence

$$\displaystyle{V (x,y) = x^{2}y + 3xy^{2} - y^{3} + C,}$$

and a general solution to the studied ODE is defined implicitly by the equation

$$\displaystyle{x^{2}y + 3xy^{2} - y^{3} = C.}$$

 □ 

Example 2.20.

Solve the differential equation

$$\displaystyle{2x\left (1 + \sqrt{x^{2 } - y}\right )dx -\sqrt{x^{2 } - y}dy = 0.}$$

Observe that

$$\displaystyle{ \frac{\partial } {\partial y}[2x(1 + \sqrt{x^{2 } - y})] = \frac{\partial } {\partial x}(-\sqrt{x^{2 } - y}) = - \frac{x} {\sqrt{x^{2 } - y}},}$$

and hence we deal with the exact differential equation. We have

$$\displaystyle{ \frac{\partial V } {\partial x} = 2x(1 + \sqrt{x^{2 } - y}),\quad \frac{\partial V } {\partial y} = -\sqrt{x^{2 } - y}, }$$

(*)

and integration of the first equation yields

$$\displaystyle{V (x,y) =\int (2x + 2x\sqrt{x^{2 } - y})dx = x^{2} + \frac{2} {3}(x^{2} - y)^{\frac{3} {2} } +\uppsi (y).}$$

Substitution of V (x, y) into the second equation of (*) gives

$$\displaystyle{ \frac{\partial } {\partial y}[x^{2} + \frac{2} {3}(x^{2} - y)^{\frac{3} {2} } +\uppsi (y)] = -\sqrt{x^{2 } - y},}$$

or equivalently

$$\displaystyle{-\sqrt{x^{2 } - y} + \frac{d\uppsi } {dy} = -\sqrt{x^{2 } - y},}$$

which means that

$$\displaystyle{\uppsi (y) = C.}$$

Finally, we have

$$\displaystyle{V (x,y) = x^{2} + \frac{2} {3}(x^{2} - y)^{\frac{3} {2} },}$$

and a general solution to the studied differential equation is

$$\displaystyle{x^{2} + \frac{2} {3}(x^{2} - y)^{\frac{3} {2} } = C}$$

or

$$\displaystyle{y = x^{2} - [\frac{3} {2}(C - x^{2})]^{\frac{2} {3} }.}$$

 □ 

Example 2.21.

A mirror reflects solar radiation in a way that a light ray coming from a source 0 after the reflection is parallel to a given direction 0X, which is the rotation axis. Figure 2.2 shows a scheme of the light ray 0A coming from the light source 0, and the rectangular coordinates 0XY. Derive the mirror shape analytically.

Since A belongs to the mirror surface, the marked angles $\varphi$ before and after reflection are equal, and n(t) denotes a normal (tangent) to the curve being intersection of the mirror and surface 0XY.

Fig. 2.2

The mirror shape y(x) and light rays

Owing to the reflection principle (the angle of incidence is equal to the reflection angle) 0A = 0B, and hence

$$\displaystyle{\text{tan}\varphi = \frac{AA^{{\prime}}} {B0 + 0A^{{\prime}}} = \frac{AA^{{\prime}}} {\sqrt{(0A^{{\prime} } )^{2 } + (A^{{\prime} } A)^{2}} + 0A^{{\prime}}},}$$

or equivalently

$$\displaystyle{\frac{dy} {dx} = \frac{y} {x + \sqrt{x^{2 } + y^{2}}}.}$$

We may rewrite the latter equation in the following way

$$\displaystyle{xdx + ydy = \sqrt{x^{2 } + y^{2}}dx,}$$

because

$$\displaystyle{dy = \frac{y(x -\sqrt{x^{2 } + y^{2}})} {(x + \sqrt{x^{2 } + y^{2}})(x -\sqrt{x^{2 } + y^{2}})}dx = \frac{x -\sqrt{x^{2 } + y^{2}}} {-y} dx.}$$

Applying the integrating multiplier

$$\displaystyle{m(x,y) = \frac{1} {\sqrt{x^{2 } + y^{2}}},}$$

we get

$$\displaystyle{\frac{xdx + ydy} {\sqrt{x^{2 } + y^{2}}} - dx = 0}$$

or equivalently

$$\displaystyle{ \frac{d(x^{2} + y^{2})} {2\sqrt{x^{2 } + y^{2}}} - dx = 0.}$$

It means that

$$\displaystyle{\sqrt{x^{2 } + y^{2}} = x + C,}$$

which allows to find the mirror surface as a paraboloid that intersects with the surface 0XY yielding a parabola governed by the equation

$$\displaystyle{y^{2} = 2Cx + C^{2}.}$$

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Example 2.22.

Solve the differential equation

$$\displaystyle{ydx - (x + x^{2} + y^{2})dy = 0}$$

assuming the integrating multiplier m = m(r(x, y)), where $r(x,y) = x^{2} + y^{2}$.

We apply formula (2.55) directly, and we get

$$\displaystyle{ \frac{1} {m} \frac{dm} {dr} = \frac{1 + 1 + 2x} {-2(x + x^{2} + y^{2})x - 2y^{2}} = \frac{2(1 + x)} {-2(1 + x)(x^{2} + y^{2})} = -\frac{1} {r}.}$$

Therefore, the following differential equation is obtained

$$\displaystyle{\frac{dm} {m} + \frac{dr} {r} = 0,}$$

which yields

$$\displaystyle{m(x,y) = \frac{1} {r(x,y)} = \frac{1} {x^{2} + y^{2}}.}$$

Now, we multiply by m the studied differential equation to get

$$\displaystyle{ \frac{ydx} {x^{2} + y^{2}} -\left ( \frac{x} {x^{2} + y^{2}} + 1\right )dy = 0,}$$

which is an exact differential equation, i.e.

$$\displaystyle{\frac{\partial V } {\partial x} = \frac{y} {x^{2} + y^{2}},\quad \frac{\partial V } {\partial y} = -\left ( \frac{x} {x^{2} + y^{2}} + 1\right ).}$$

Integration of the first equation in the above gives

$$\displaystyle{V (x,y) =\int \frac{y} {x^{2} + y^{2}}dx = \text{arctan}\frac{x} {y} +\uppsi (y),}$$

and hence

$$\displaystyle{\frac{\partial V } {\partial y} = \frac{\partial } {\partial y}\left (\text{arctan}\frac{x} {y} +\uppsi (y)\right ) = - \frac{x} {x^{2} + y^{2}} - 1.}$$

It means that

$$\displaystyle{- \frac{1} {1 + \frac{x^{2}} {y^{2}} } \frac{x} {y^{2}} + \frac{d\uppsi } {dy} = - \frac{x} {x^{2} + y^{2}} - 1,}$$

and finally

$$\displaystyle{ \frac{d\uppsi } {dy} = -1,\quad \uppsi (y) = -y + C_{1},}$$

and

$$\displaystyle{V (x,y) = \text{arctan}\frac{x} {y} - y + C_{1}.}$$

We have the following solutions: one given explicitly y = 0, and other given implicitly

$$\displaystyle{\text{arctan}\frac{x} {y} - y = C.}$$

2.6 Implicit Differential Equations Not Solved with Respect to a Derivative

We consider here the differential equation (2.2), which cannot be solved with respect to $\frac{dy} {dt}$, i.e. we cannot reduce the problem to that of Eq. (2.3). It may happen, however, that Eq. (2.2) can be solved with respect to either x or y. In what follows we describe briefly the method of the parameter introduction yielding a solution in the latter case. Let

$$\displaystyle{ y = f(x,y^{{\prime}}),\quad y^{{\prime}}\equiv \frac{dy} {dx} = p, }$$

(2.56)

where p is the introduced parameter. The full differential of $y = f(x,y^{{\prime}})$ follows

$$\displaystyle{ pdx = \frac{\partial f} {\partial x}dx + \frac{\partial (x,p)} {\partial p} dp. }$$

(2.57)

It means that we have got the exact differential equation form (2.44), where

$$\displaystyle{ M(x,y) = \frac{\partial f} {\partial x},\quad N(x,y) = \frac{\partial (x,p)} {\partial p}. }$$

(2.58)

In the previous section supplemented by many examples we have described various methods yielding solutions to Eq. (2.56). Namely, we can take

$$\displaystyle{ x =\uppsi (p,c),\quad y = f(x,p), }$$

(2.59)

where $x =\uppsi (p,c)$ is the implicit form of solution governed by Eq. (2.56).

Theorem 2.2.

Suppose that the function f(x,y,y ^′ ) in a neighbourhood of the point $(x_{0},y_{0},y_{0}^{{\prime}})$ , where $y_{0}^{{\prime}}$ is one of the roots of the equation $f(x_{0},y_{0},y_{0}^{{\prime}}) = 0$ , is continuous regarding x and it is continuously differentiable with respect to y, y ^′ , and $\frac{\partial f} {\partial y^{{\prime}}}(x_{0},y_{0},y_{0}^{{\prime}})\neq 0$ . Then there exists a unique solution y = ψ ^′ (x) of the Cauchy problem $f(x,y,y^{{\prime}}) = 0$, $y(x_{0}) = y_{0}$ defined in a satisfactorily close neighbourhood of the point x ₀ , where $\psi ^{{\prime}}(x_{0}) = y_{0}^{{\prime}}$ .

Recall that the uniqueness of problem of Eq. (2.2) means that the point $(x_{0},y_{0})$ is a point of the solution uniqueness, i.e. there are no other integral curves of (2.2) which pass through the point $(x_{0},y_{0})$ and have the same slope in this point. Otherwise, the solution uniqueness is violated.

Theorem 2.2 yields sufficient conditions of a solution existence and uniqueness for Eq. (2.2).

Assuming that the function $f(x,y,y^{{\prime}})$ is continuous with respect to x and continuously differentiable with respect to y and y ^′, then a possible set of singular points is defined via the following system of algebraic equations

$$\displaystyle\begin{array}{rcl} f(x,y,y^{{\prime}})& =\,& 0, \\ \frac{\partial f} {\partial y^{{\prime}}}(x,y,y^{{\prime}})& =\,& 0.{}\end{array}$$

(2.60)

It is required, while solving Eq. (2.2) to find singular solution, i.e. we remove y ^′ from Eq. (2.60) and we get a so-called discriminant-type curve. Each branch of this curve should be verified if it is a solution to Eq. (2.2). Assuming a positive reply, our next step consists of checking if its points correspond to the solution non-uniqueness.

The method of parameter introduction can be directly applied either to the so-called Claurait equation

$$\displaystyle{ y = xy^{{\prime}} +\uppsi (y^{{\prime}}), }$$

(2.61)

or to the so-called Lagrange equation

$$\displaystyle{ y = x\varphi (y^{{\prime}}) +\uppsi (y^{{\prime}}). }$$

(2.62)

Example 2.23.

Solve the following Claurait equation

$$\displaystyle{\sqrt{(y^{{\prime} } )^{2 } + 1} + xy^{{\prime}}- y = 0.}$$

We introduce p = y ^′ to get

$$\displaystyle{y = xp + \sqrt{1 + p^{2}}.}$$

Differentiation of the last equation with respect to x yields

$$\displaystyle{\frac{dy} {dx} = p + x\frac{dp} {dx} + \frac{p\frac{dp} {dx}} {\sqrt{1 + p^{2}}},}$$

and hence

$$\displaystyle{\left (x + \frac{p} {\sqrt{1 + p^{2}}}\right )\frac{dp} {dx} = 0.}$$

It means that either

$$\displaystyle{x = - \frac{p} {\sqrt{1 + p^{2}}}.}$$

or

$$\displaystyle{p = C.}$$

A solution to the problem is as follows:

$$\displaystyle{y = Cx + \sqrt{1 + C^{2}}}$$

or equivalently

$$\displaystyle\begin{array}{rcl} x& =\,& - \frac{p} {\sqrt{1 + p^{2}}}, {}\\ y& =\,& px + \sqrt{1 + p^{2}}. {}\\ \end{array}$$

 □ 

Example 2.24.

Solve the following Lagrange equation

$$\displaystyle{y^{{\prime}} + y = x(y^{{\prime}})^{2}.}$$

It is easily solved with respect to y, i.e.

$$\displaystyle{y = x(y^{{\prime}})^{2} - y^{{\prime}}}$$

or equivalently

$$\displaystyle{y = xp^{2} - p,}$$

where p = y ^′. Differentiation of this algebraic equation yields

$$\displaystyle{p \equiv \frac{dy} {dx} = p^{2} + 2px\frac{dp} {dx} -\frac{dp} {dx},}$$

or equivalently

$$\displaystyle\begin{array}{rcl} & p(p - 1)\frac{dx} {dp} = (1 - 2px),& {}\\ & \frac{dx} {dp} + \frac{2x} {p-1} = \frac{1} {p(p-1)}. & {}\\ \end{array}$$

In other words, the problem has been reduced to a linear differential equation with the following solution

$$\displaystyle{x = \frac{p -\ln p + C} {(p - 1)^{2}}.}$$

 □ 

Example 2.25.

Derive an equation governing a family of equipotential curves of the electric field generated by a dipole. Recall that the equipotential curves are orthogonal to force curves of the electric field (see Example 2.3).

As it has been shown previously in Example 2.3, we have

$$\displaystyle{\left (\frac{x - a} {r_{2}^{3}} -\frac{x + a} {r_{1}^{3}} \right )\frac{dy} {dx} -\left ( \frac{1} {r_{2}^{3}} - \frac{1} {r_{1}^{3}}\right ) = 0,}$$

where

$$\displaystyle{ r_{1}^{2} = (x + a)^{2} + y^{2},\quad r_{ 2}^{2} = (x - a)^{2} + y^{2}. }$$

(*)

We may generalize the studied case in Example 2.3. Namely, we began with the algebraic problem governed by the following equation

$$\displaystyle{F(x,y,a) = 0,}$$

where

$$\displaystyle{F(x,y,a) = \frac{x + a} {\sqrt{(x + a)^{2 } + y^{2}}} - \frac{x - a} {\sqrt{(x - a)^{2 } + y^{2}}} - C.}$$

For a given C, we have a family of one parameter curves. In what follows we define another family of the isogonal curves, which interset the first family curves with the same angle $\varphi$, for $\varphi =\pi /2$ we say that both trajectories (curves) are orthogonal.

We differentiate the algebraic equation to get

$$\displaystyle{\frac{dF} {dx} \equiv \frac{\partial F(x,y,a)} {\partial x} + \frac{\partial F(x,y,a)} {\partial y} \frac{dy} {dx} = 0.}$$

We may also exclude the parameter a using the equation F = 0. In our case we have

$$\displaystyle{\frac{\partial F} {\partial x} = \left ( \frac{1} {r_{2}^{3}} - \frac{1} {r_{1}^{3}}\right )y,\quad \frac{\partial F} {\partial y} = \frac{x - a} {r_{2}^{3}} -\frac{x - a} {r_{1}^{3}}.}$$

In Fig. 2.3 two curves belonging to both families are shown intersecting in the point A = A(x, y).

Fig. 2.3

Two curves 1 and 2 intersecting in point A

The angle between two curves at point A is $\varphi$ (known), which is given by the formula

$$\displaystyle{\pi =\alpha +\varphi +\pi -\beta.}$$

Therefore, we get

$$\displaystyle{\text{tan}\beta = \text{tan}(\alpha +\varphi ) = \frac{\text{tan}\alpha + \text{tan}\varphi } {1 -\text{tan}\alpha \text{tan}\varphi }.}$$

We apply the following notation $\text{tan}\alpha = y^{{\prime}}$, $\text{tan}\beta = y_{{\ast}}^{{\prime}}$, $\text{tan}\varphi = m$, and hence

$$\displaystyle{y_{{\ast}}^{{\prime}} = \frac{y^{{\prime}} + m} {1 - my^{{\prime}}}.}$$

In a case of orthogonal trajectories we have $\varphi =\pi /2$, and therefore

$$\displaystyle\begin{array}{rcl} & \text{tan}\beta = \frac{\text{tan}\alpha +\text{tan} \frac{\pi } {2} } {1-\text{tan}\alpha \text{tan} \frac{\pi } {2} } & {}\\ & =\lim _{\varphi \rightarrow \frac{\pi }{ 2} } \frac{1+\frac{\text{tan}\alpha }{\text{tan}\varphi }} { \frac{1} {\text{tan}\varphi }-\text{tan}\alpha } = - \frac{1} {\text{tan}\alpha }& {}\\ \end{array}$$

or equivalently

$$\displaystyle{y_{{\ast}}^{{\prime}} = -\frac{1} {y^{{\prime}}}.}$$

The so far consideration implies a simple recipe. In order to find a differential equations of the family of isogonal trajectories to the trajectories (curves) governed by equation F(x, y, a) = 0, we need to substitute the term $y^{{\prime}} = \frac{dy} {dx}$ standing in equation $\partial F/\partial x + \partial F/\partial yy^{{\prime}} = 0$, by the term $y_{{\ast}}^{{\prime}}$. In a case for $\varphi = \frac{\pi } {2}$ (orthogonal trajectories) we substitute y ^′ by $-\frac{1} {y^{{\prime}}} = -\frac{dx} {dy}$.

In the studied case, using the so far described orthogonality property we obtain the following differential equation

$$\displaystyle{\left (\frac{x - a} {r_{2}^{3}} -\frac{x + a} {r_{1}^{3}} \right )\left (- \frac{1} {\frac{dy} {dx}}\right ) -\left ( \frac{1} {r_{2}^{3}} - \frac{1} {r_{1}^{3}}\right )y = 0,}$$

or equivalently

$$\displaystyle{(x - a)r_{1}^{3} - (x + a)r_{ 2}^{3} + y(r_{ 1}^{3} - r_{ 2}^{3})\frac{dy} {dx} = 0.}$$

From (*) we get

$$\displaystyle\begin{array}{rcl} r_{1}dr_{1}& =& (x + a)dx + ydy, {}\\ r_{2}dr_{2}& =& (x - a)dx + ydy, {}\\ \end{array}$$

therefore the problem is reduced to the following differential equation

$$\displaystyle{r_{1}^{3}r_{ 2}dr_{2} = r_{2}^{3}r_{ 1}dr_{1},}$$

which yields the following solution

$$\displaystyle{ \frac{1} {r_{2}} - \frac{1} {r_{1}} = C,}$$

and hence

$$\displaystyle{ \frac{1} {\sqrt{(x - a)^{2 } + y^{2}}} - \frac{1} {\sqrt{(x + a)^{2 } + y^{2}}} = C.}$$

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