# 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.