Show that the differential equation `2xydy/dx=x^2+3y^2` is homogeneous and solve it.

#### Solution

The given differential equation can be expressed as

`dy/dx=(x^2+3y^2)/(2xy) .....(i)`

`Let F(x, y)=(x^2+3y^2)/(2xy)`

Now,

`F(λx, λy)=((λx)^2+3(λy)^2)/(2(λx)(λy))=(λ^2(x^2+3y^2))/(λ^2(2xy))=λ^0F(x, y)`

Therefore, F(*x*, *y*) is a homogenous function of degree zero. So, the given differential equation is a homogenous differential equation.

Let *y* = *vx .....*(ii)

Differentiating (ii) w.r.t. *x*, we get

`dy/dx=v+x(dv)/dx`

Substituting the value of y and dy/dx in (i), we get

`v+x(dv)/dx=(1+3v^2)/(2v)`

`⇒x(dv)/dx=(1+3v^2)/(2v)−v`

` ⇒x(dv)/dx=(1+3v^2−2v^2)/(2v)`

`⇒x(dv)/dx=(1+v^2)/(2v)`

`⇒(2v)/(1+v^2)dv=dx/x .....(ii)`

Integrating both side of (iii), we get

`∫(2v)/(1+v^2)dv=∫dx/x`

Putting `1+v^2=t`

⇒2vdv=dt

`∴∫dt/t=∫dx/x`

⇒log|t|=log|x| +log|C_{1}|

⇒log∣t/x∣=log|C_{1}|

`⇒t/x=±C_1`

`⇒(1+v^2)/x=±C_1`

`⇒(1+y^2/x^2)/x=±C_1`

x^{2}+y^{2}=Cx^{3}