Question

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

Answer 1

Notice that

$\dfrac{\partial(x+y)^2}{\partial y}=2(x+y)=2x+2y$

$\dfrac{\partial(2xy+x^2-8)}{\partial x}=2y+2x$

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

$\dfrac{\partial f}{\partial x}\,\mathrm dx+\dfrac{\partial f}{\partial y}\,\mathrm dy=0$

$\dfrac{\partial f}{\partial x}=(x+y)^2=x^2+2xy+y^2\implies f(x,y)=\dfrac{x^3}3+x^2y+xy^2+g(y)$

$\dfrac{\partial f}{\partial y}=2xy+x^2-9=x^2+2xy+\dfrac{\mathrm dg}{\mathrm dy}\implies\dfrac{\mathrm dg}{\mathrm dy}=-9\implies g(y)=-9y+C$

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

so that the solution to the ODE is

$\dfrac{x^3}3+x^2y+xy^2-9y=C$

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

$\dfrac13+1+1-9=C\implies C=-\dfrac{20}3$

so that the particular solution is

$\boxed{\dfrac{x^3}3+x^2y+xy^2-9y=-\dfrac{20}3}$