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

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

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

$\Psi_y=2xy+x^2-2=(x+y)^2+f'(y)$
$\implies2xy+x^2-2=x^2+2xy+y^2+f'(y)$
$f'(y)=-2-y^2\implies f(y)=-2y-\dfrac{y^3}3+C$

So the solution to this ODE is

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

i.e.


$\dfrac{(x+y)^3}3-2y-\dfrac{y^3}3=C$