Homogeneous Differential Equations

Suppose we have a differential equation of the form

$ \displaystyle \frac{dy}{dx} = f(x,y)$ . . . . . (1)

where f(x,y) is a function of x and y and is of the form

$\displaystyle F(\frac{y}{x}) \; or \; F(\frac{x}{y}) $

These equations are solved by putting y = vx, where v ≡ v (x) is a function of x.

Illustration : Solve (x3 − 2y3)dx + 3xy2 dy = 0

Solution : Here

$ \displaystyle \frac{dy}{dx} = -\frac{(x^3-2y^3)}{3xy^2} $

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

Let y = vx

$ \displaystyle \frac{dy}{dx} = v + x\frac{dv}{dx} $

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

$ \displaystyle x\frac{dv}{dx} = -\frac{v}{3}-\frac{1}{3v^2} $

$ \displaystyle x\frac{dv}{dx} = -\frac{v^3 + 1}{3v^2} $

$ \displaystyle \frac{3v^2}{v^3 +1}dv = -\frac{dx}{x} $

Integrating ,

$ \displaystyle \int \frac{3v^2}{v^3 +1}dv = -\int \frac{dx}{x} $

⇒ ln (v3 + 1) = −ln x + ln c

⇒ ln (v3 + 1) + ln x = ln c

⇒ ln x (v3 + 1) = ln c

x (v3 + 1) = c

$ \displaystyle x(\frac{y^3}{x^3} +1) = c$

$ \displaystyle y^3 + x^3 = c x^2 $

Illustration : Solve

$ \displaystyle (1-2 e^{x/y})dx -2 e^{x/y} (1-\frac{x}{y})dy = 0 $

Solution:

$ \displaystyle (1-2 e^{x/y})dx -2 e^{x/y} (1-\frac{x}{y})dy = 0 $

Putting x = vy, we get dx = vdy + ydv

Therefore (1 − 2ev) (vdy + ydv) – 2ev (1−v) dy = 0

⇒ (v − 2ev) dy + (1 − 2ev)y dv = 0

$ \displaystyle \frac{dy}{y} + \frac{1-2e^v}{v-2e^v}dv = 0 $

Integrating, we get

$ \displaystyle ln y + ln(v-2e^v) = lnc $

$ \displaystyle ln y (v-2e^v) = lnc $

$ \displaystyle y ( \frac{x}{y} -2e^{x/y}) = c $

⇒ x − 2 y ex/y = c

Equations Reducible to the Homogenous form

Equations of the form $ \displaystyle \frac{dy}{dx}= \frac{ax+by+c}{AX +BY +C} $ (aB ≠ Ab)

can be reduced to a homogenous form by changing the variables x, y to X, Y by writing x = X + h and y = Y + k; where h, k are constants to be chosen so as to make the given equation homogenous. We have

$ \displaystyle \frac{dy}{dx}= \frac{d(Y+k)}{d(X+h)} = \frac{dY}{dX} $

Hence equation becomes

$ \displaystyle \frac{dY}{dX}= \frac{aX + bY + (ah+bk+c)}{AX +BY + (Ah +Bk + C)} $

Let h and k be so chosen as to satisfy the relation ah + bk + c = 0 and Ah + Bk + C =0

These give

$ \displaystyle h = \frac{bC-Bc}{aB-Ab} \; , \; k= \frac{Ac-aC}{aB-Ab} $

which are meaningful except when aB = Ab

$ \displaystyle \frac{dY}{dX}= \frac{aX + bY}{AX + BY} $

can now be solved by means of the substitution Y = VX

In case aB = Ab, we write ax + by = t
This reduces the differential equation to the separable variable type.

Illustration : Solve :

$ \displaystyle \frac{dy}{dx}= \frac{x+2y+3}{2x+3y+4} $

Solution : Put x = X + h , y = Y + k

We have

$ \displaystyle \frac{dY}{dX}= \frac{X + 2Y + (h+2k+3)}{2X + 3Y +(2h+2k+ 4)} $

To determine h and k , we write

h + 2k + 3 = 0 , 2h + 3k + 4 = 0

=> h = 1 , k = −2

so that

$ \displaystyle \frac{dY}{dX} = \frac{X+2Y}{2X+3Y} $

Putting Y = VX , we get

$ \displaystyle \frac{dY}{dX}= V + X \frac{dV}{dX} $

$ \displaystyle V + X \frac{dV}{dX} = \frac{1+2V}{2+3V} $

$ \displaystyle X \frac{dV}{dX} = \frac{1+2V}{2+3V} -V $

$ \displaystyle \frac{2+3V}{3V^2-1}dV = -\frac{dX}{X} $

$ \displaystyle [\frac{2+\sqrt{3}}{2(\sqrt{3}V-1)} -\frac{2-\sqrt{3}}{2(\sqrt{3}V+1)}]dV = -\frac{dX}{X} $

$ \displaystyle \frac{2+\sqrt{3}}{2\sqrt{3}}log(\sqrt{3}V-1) – \frac{2-\sqrt{3}}{2\sqrt{3}}log(\sqrt{3}V+1) = -logX +c $

$ \displaystyle \frac{2+\sqrt{3}}{2\sqrt{3}}log(\sqrt{3}V-1) – \frac{2-\sqrt{3}}{2\sqrt{3}}log(\sqrt{3}V+1) + logX = c $

, where X = x − 1 , Y = y + 2

Illustration: Solve $\large \frac{dy}{dx} = \frac{4x + 6y -5}{6x + 9y + 7}$

Solution: Here aB = A b ( since 4× 9 = 6×6)

Putting 4x + 6y = t and $ 4 + 6\frac{dy}{dx} = \frac{dt}{dx}$

$\large \frac{dy}{dx} = \frac{1}{6} (\frac{dt}{dx} – 4)$

The  differential equation becomes ,

$\large  \frac{1}{6} (\frac{dt}{dx} – 4) = \frac{t-5}{\frac{3}{2}t + 7} = \frac{2t-10}{3t+14}$

$\large   (\frac{dt}{dx} – 4) = \frac{12t-60}{3t+14}$

$\large   \frac{dt}{dx}  = \frac{12t-60}{3t+14} + 4 = \frac{24t -4}{3t+14}$

$\large \int \frac{24t -4}{3t+14} dt = \int dx$

$\large \frac{1}{8} \int dt + \frac{29}{8} \int \frac{dt}{6t-1} = \int dx $

$\large \frac{1}{8} t + \frac{29}{8} \frac{1}{6}ln |6t-1| = x + c $

$\large \frac{1}{8} (4x+6y) + \frac{29}{48} ln |24x + 36y -1| = x + c $ , is the required solution.

Exercise : Solve the following differential equations

(i) $\large \frac{dy}{dx} = \frac{y}{x} + tan\frac{y}{x} $

(ii) $\large \frac{dy}{dx} = \frac{ax + by – a}{bx + ay -b} $

(iii) $\large (2x + y + 1)dx + (4x + 2y – 1)dy = 0 $

(iv) $\large \frac{dy}{dx} = \frac{x^2 + y^2 – 2x – 2y + 2}{x y – x -y + 1} $

Next Page→

← Back Page

Leave a Reply