Question

Please help me solve this DE problem: `(4x^2-y^2)dx-xydy=0`

Answer

Rewrite equation:

`4x^2-y^2-xyy'=0`

Let `y=ux` , then `y'=u'x+u`.

Plugging these values into initial equation we get that

`4x^2-u^2x^2-x*ux*(u'x+u)=0`

`4-u^2-u(u'x+u)=0`

`4-2u^2=u'ux`

`(4-2u^2)=(du)/(dx)ux`

`(dx)/x=u/(4-2u^2)du`

Integrate both sides of the equation:

`ln(|x|)=-1/4 ln(|4-2u^2|)+A`

`-4ln(|x|)=ln(|4-2u^2|)-4A`

Change constant:

`ln(1/x^4)=ln(|4-2u^2|)+ln(B)`

`ln(1/x^4)=ln(B|4-2u^2|)`

`1/x^4=B(4-2u^2)`

Recall that `y=ux` or `u=y/x`:

`1/x^4=B(4-2y^2/x^2)`

`1=B(4x^4-2x^2y^2)`

`-1/(2B)=2x^4-x^2y^2`

Change constant, let `C=-1/(2B)`:

`C=x^2(2x^2-y^2)`

`x^2(2x^2-y^2)=C`