Question

Find the general solution for the following homogeneous equations. PLEASE SHOW SOLUTIONS

1) (xy+y2)dx−x2dy=0 (hint: let u=y/x, so y=ux, dy=xdu+udx))

2) 2x2ydx=(3x3+y3)dy. (hint: let v=x/y, so x=vy, dx=vdy+ydv)

Answer 1

Follow the hints.

1. Let
`u=\frac yx`, so that
`y=ux` and
`\mathrm dy=x,\mathrm du+u\,\mathrm dx`. Substituting into

`(xy+y^2)\,\mathrm dx-x^2\,\mathrm dy=0`

gives

`(ux^2+u^2x^2)\,\mathrm dx-x^2(x\,\mathrm du+u\,\mathrm dx)=0`

`u^2x^2\,\mathrm dx-x^3\,\mathrm du=0`

and the remaining ODE is separable:

`x^3\,\mathrm du=u^2x^2\,\mathrm dx\implies(\mathrm du)/(u^2)=\frac{\mathrm dx}x`

Integrate both sides to get

`-\frac1u=\ln|x|+C`

`-\frac xy=\ln|x|+C`

`\boxed{y=\frac xCx-\ln}`

2.
`Let [tex]v=\frac xy`, so that
`x=vy` and
`\mathrm dx=v\,\mathrm dy+y\,\mathrm dv`. Then

`2x^2y\,\mathrm dx=(3x^3+y^3)\,\mathrm dy`

becomes

`2v^2y^3(v\,\mathrm dy+y\,\mathrm dv)=(3v^3y^3+y^3)\,\mathrm dy`

`2v^3y^3\,\mathrm dy+2v^2y^4\,\mathrm dv=(3v^3y^3+y^3)\,\mathrm dy`

`2v^2y^4\,\mathrm dv=(v^3y^3+y^3)\,\mathrm dy`

which is separable as

`(2v^2)/(v^3+1)\,\mathrm dv=\frac{\mathrm dy}y`

Integrating both sides gives

`\frac23\ln|v^3+1|=\ln|y|+C`

`\ln|v^3+1|=\frac32\ln|y|+C`

`v^3+1=Cy^(3/2)`

`v=\sqrt[3]{Cy^(3/2)-1}`

`\frac xy=\sqrt[3]{Cy^(3/2)-1}`

`\boxed{x=y\sqrt[3]{Cy^(3/2)-1}}`