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

Notice that[tex]\dfrac{\partial(x+y)^2}{\partial y}=2(x+y)=2x+2y[/tex][tex]\dfrac{\partial(2xy+x^2-8)}{\partial x}=2y+2x[/tex]so the ODE is exact, and we can look for a solution of the form [tex]f(x,y)=C[/tex]. Differentiating both sides gives[tex]\dfrac{\partial f}{\partial x}\,\mathrm dx+\dfrac{\partial f}{\partial y}\,\mathrm dy=0[/tex][tex]\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)[/tex][tex]\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[/tex][tex]\implies f(x,y)=\dfrac{x^3}3+x^2y+xy^2-9y+C=C[/tex]so that the solution to the ODE is[tex]\dfrac{x^3}3+x^2y+xy^2-9y=C[/tex]Given that [tex]y(1)=1[/tex], we have[tex]\dfrac13+1+1-9=C\implies C=-\dfrac{20}3[/tex]so that the particular solution is[tex]\boxed{\dfrac{x^3}3+x^2y+xy^2-9y=-\dfrac{20}3}[/tex]