Solution of Homogeneous Differential Equation

Here you will learn how to find solution of homogeneous differential equation of first order first degree with examples.

Let’s begin –

Solution of Homogeneous Differential Equation

If a first degree first order differential equation is expressible in the form

$dy\over dx$ = $f(x, y)\over g(x, y)$

where f(x, y) and g(x, y) are homogeneous function of the same degree, then it is called a homogeneous differential equation,

Such type of equation can be reduced to variable seperable form by the substitution y = vx. Process is shown in the algorithm below.

Algorithm :

1). Put the differential equation in the form

$dy\over dx$ = $f(x, y)\over g(x, y)$

2). Put y = vx and $dy\over dx$ = v + x$dy\over dx$ in the equation in step 1 and cancel out x from the right hand side.

3). Shift v on RHS and seperate the variables v and x.

4). Integrate both sides to obtain the solution in terms of v and x.

5). Replace v by $y\over x$ in the solution obtained in step 4 to obtain the solution in terms of x and y.

Example : Solve the differential equation $x^2$ dy + y(x + y) dx = 0.

Solution : The given differential equation is

$x^2$ dy + y(x + y) dx = 0

$\implies$ $x^2$ dy = -y(x + y) dx

$\implies$ $dy\over dx$ = -($xy + y^2\over x^2$)                      …………………(i)

Since each of the functions xy + $y^2$ and $x^2$ is a homogeneous function of degree 2.

Therefore, equation (i) is a homogeneous differential equation.

Putting y = vx and $dy\over dx$ = v + x$dv\over dx$ in (i), we get

v + x$dv\over dx$ = -($vx^2 + v^2x^2\over x^2$)

v + x$dv\over dx$ = -($v + v^2$)

x$dv\over dx$ = -($2v + v^2$)

xdv = -($v^2 + 2v$)dx

By Seperating the variable,

$1\over v^2 + 2v$ dv = $-dx\over x$

Integrating both sides.

$\implies$ $\int$ $1\over v^2 + 2v$ dv = $\int$ $-1\over x$ dx

$\int$ $1\over v^2 + 2v + 1- 1$ dv = $\int$ $-1\over x$ dx

$\int$ $1\over (v + 1)^2 – 1^2$ dv = $\int$ $-1\over x$ dx

$1\over 2$ $log{{v+1 – 1}\over {v+1+1}}$ = -log x + log C

$1\over 2$ $log{{v}\over {v+2}}$ = -log x + log C

$log{{v}\over {v+2}}$ + 2log x =  2log C

$log|{vx^2\over v+2}|$ = log k

Put v = y/x

k = $x^2y\over y + 2x$, which is the solution of differential equation.