Question

Solve the given initial-value problem. (y2 cos x − 3x2y − 4x) dx + (2y sin x − x3 + ln y) dy = 0, y(0) = e

Answer 1

$\underbrace{(y^2\cos x-3x^2y-4x)}_M\,\mathrm dx+\underbrace{(2y\sin x-x^3+\ln y)}_N\,\mathrm dy=0$

The ODE is exact if $M_y=N_x$. We have

$M_y=2y\cos x-3x^2$
$N_x=2y\cos x-3x^2$

and so the equation is indeed exact. So we're looking for a solution of the form $\Psi(x,y)=C$, noting that the total differential for this solution is

$\Psi_x\,\mathrm dx+\Psi_y\,\mathrm dy=0$

which corresponds to

$\begin{cases}\Psi_x=M\\\Psi_y=N\end{cases}$


In the first PDE, we can integrate both sides with respect to $x$ to get

$\displaystyle\int\Psi_x\,\mathrm dx=\int(y^2\cos x-3x^2y-4x)\,\mathrm dx$
$\implies\Psi(x,y)=y^2\sin x-x^3y-2x^2+f(y)$

where $f(y)$ is assumed to be a function of $y$ alone. Then differentiating with respect to $y$ gives us

$\Psi_y=2y\sin x-x^3+f'(y)=2y\sin x-x^3+\ln y$
$\implies f'(y)=\ln y$
$\implies f(y)=y\ln y-y+C$

So the general solution is

$\Psi(x,y)=y^2\sin x-x^3y-2x^2+y\ln y-y+C=C$

or simply

$y^2\sin x-x^3y-2x^2+y\ln y-y=C$

Given that $y(0)=e$, we have

$e^2\sin0-0^3\cdot e-2\cdot0^2+e\ln e-e=C$
$\implies C=0$

so the particular solution to the IVP is

$y^2\sin x-x^3y-2x^2+y\ln y-y=0$