Question

(X^2+y^2+x)dx+xydy=0
Solve for general solution

Answer 1

Check if the equation is exact, which happens for ODEs of the form

`M(x,y)\,\mathrm dx+N(x,y)\,\mathrm dy=0`

if
`(\partial M)/(\partial y)=(\partial N)/(\partial x)`.

We have

`M(x,y)=x^2+y^2+x\implies(\partial M)/(\partial y)=2y`

`N(x,y)=xy\implies(\partial N)/(\partial x)=y`

so the ODE is not quite exact, but we can find an integrating factor
`\mu(x,y)` so that

`\mu(x,y)M(x,y)\,\mathrm dx+\mu(x,y)N(x,y)\,\mathrm dy=0`

is exact, which would require

`(\partial(\mu M))/(\partial y)=(\partial(\mu N))/(\partial x)\implies (\partial\mu)/(\partial y)M+\mu(\partial M)/(\partial y)=(\partial\mu)/(\partial x)N+\mu(\partial N)/(\partial x)`

`\implies\mu\left((\partial N)/(\partial x)-(\partial M)/(\partial y)\right)=M(\partial\mu)/(\partial y)-N(\partial\mu)/(\partial x)`

Notice that

`(\partial N)/(\partial x)-(\partial M)/(\partial y)=y-2y=-y`

is independent of x, and dividing this by
`N(x,y)=xy` gives an expression independent of y. If we assume
`\mu=\mu(x)` is a function of x alone, then
`(\partial\mu)/(\partial y)=0`, and the partial differential equation above gives

`-\mu y=-xy(\mathrm d\mu)/(\mathrm dx)`

which is separable and we can solve for
`\mu` easily.

`-\mu=-x(\mathrm d\mu)/(\mathrm dx)`

`\frac{\mathrm d\mu}\mu=\frac{\mathrm dx}x`

`\ln|\mu|=\ln|x|`

`\implies \mu=x`

So, multiply the original ODE by x on both sides:

`(x^3+xy^2+x^2)\,\mathrm dx+x^2y\,\mathrm dy=0`

Now

`(\partial(x^3+xy^2+x^2))/(\partial y)=2xy`

`(\partial(x^2y))/(\partial x)=2xy`

so the modified ODE is exact.

Now we look for a solution of the form
`F(x,y)=C`, with differential

`\mathrm dF=(\partial F)/(\partial x)\,\mathrm dx+(\partial F)/(\partial y)\,\mathrm dy=0`

The solution F satisfies

`(\partial F)/(\partial x)=x^3+xy^2+x^2`

`(\partial F)/(\partial y)=x^2y`

Integrating both sides of the first equation with respect to x gives

`F(x,y)=\frac{x^4}4+\frac{x^2y^2}2+\frac{x^3}3+f(y)`

Differentiating both sides with respect to y gives

`(\partial F)/(\partial y)=x^2y+(\mathrm df)/(\mathrm dy)=x^2y`

`\implies(\mathrm df)/(\mathrm dy)=0\implies f(y)=C`

So the solution to the ODE is

`F(x,y)=C\iff \frac{x^4}4+\frac{x^2y^2}2+\frac{x^3}3+C=C`

`\implies\boxed{\frac{x^4}4+\frac{x^2y^2}2+\frac{x^3}3=C}`