Question

Solve the given initial-value problem. (x + y)^2 dx + (2xy + x^2 - 8) dy = 0, y(1) = 1

Answer 1

Notice that

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

`(\partial(2xy+x^2-8))/(\partial x)=2y+2x`

so the ODE is exact, and we can look for a solution of the form
`f(x,y)=C`. Differentiating both sides gives

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

`(\partial f)/(\partial x)=(x+y)^2=x^2+2xy+y^2\implies f(x,y)=\frac{x^3}3+x^2y+xy^2+g(y)`

`(\partial f)/(\partial y)=2xy+x^2-9=x^2+2xy+(\mathrm dg)/(\mathrm dy)\implies(\mathrm dg)/(\mathrm dy)=-9\implies g(y)=-9y+C`

`\implies f(x,y)=\frac{x^3}3+x^2y+xy^2-9y+C=C`

so that the solution to the ODE is

`\frac{x^3}3+x^2y+xy^2-9y=C`

Given that
`y(1)=1`, we have

`\frac13+1+1-9=C\implies C=-\frac{20}3`

so that the particular solution is

`\boxed{\frac{x^3}3+x^2y+xy^2-9y=-\frac{20}3}`