Question

Solve the given initial-value problem. (x + y)2 dx + (2xy + x2 − 2) dy = 0, y(1) = 1

Answer 1

Let's check if the ODE is exact. To do that, we want to show that if


`\underbrace{(x+y)^2}_M\,\mathrm dx+\underbrace{(2xy+x^2-2)}_N\,\mathrm dy=0`

then
`M_y=N_x`. We have


`M_y=2(x+y)`

`N_x=2y+2x=2(x+y)`

so the equation is indeed exact. We're looking for a solution of the form
`\Psi(x,y)=C`. Computing the total differential yields the original ODE,


`\mathrm d\Psi=\Psi_x\,\mathrm dx+\Psi_y\,\mathrm dy=0`

`\implies\begin{cases}\Psi_x=(x+y)^2\\\Psi_y=2xy+x^2-2\end{cases}`

Integrate both sides of the first PDE with respect to
`x`; then


`\displaystyle\int\Psi_x\,\mathrm dx=\int(x+y)^2\,\mathrm dx\implies\Psi(x,y)=\frac{(x+y)^3}3+f(y)`

where
`f(y)` is a function of
`y` alone. Differentiate this with respect to
`y` so that


`\Psi_y=2xy+x^2-2=(x+y)^2+f'(y)`

`\implies2xy+x^2-2=x^2+2xy+y^2+f'(y)`

`f'(y)=-2-y^2\implies f(y)=-2y-\frac{y^3}3+C`

So the solution to this ODE is


`\Psi(x,y)=\frac{(x+y)^3}3-2y-\frac{y^3}3+C=C`

i.e.



`\frac{(x+y)^3}3-2y-\frac{y^3}3=C`