Solve PDE $u_{xy}+xyu+yu_{x} + xu_{y}=0$

Question

I need help to solve differential equation $u_{xy}+xyu+yu_{x} + xu_{y}=0$.

Equation and beginning of my solution is on photo.

Here $u_{y}$ is derivative with respect to $y$ etc.

Answer 1

Now by multiplying both sides in $e^{y^2\over 2}$ we obtain$$e^{y^2\over 2}u_y+yue^{y^2\over 2}=C(y)e^{y^2-x^2\over 2}\implies\\(e^{y^2\over 2}u)_y=C(y)e^{y^2-x^2\over 2}\implies \\e^{y^2\over 2}u=e^{-x^2\over 2}\int C(y)e^{y^2\over 2}dy+D(x)$$finally$$u=e^{-{x^2\over 2}}C(y)+e^{-{y^2\over 2}}D(x)$$which can alternatively be written as$$u=e^{-x^2\over2}e^{-y^2\over2}\left(e^{y^2\over 2}C(y)+e^{x^2\over2}D(x)\right)$$Now by defining $e^{x^2\over2}D(x)\to D(x)$ and $e^{y^2\over 2}C(y)\to C(y)$ we have$$u=e^{-{x^2\over2}-{y^2\over2}}\left(D(x)+C(y)\right)$$

Answer 2

Another approach

$$ \left(\frac 1x\partial_x+1\right)\left(\frac 1y\partial_y+1\right)u = 0 $$

so now calling

$$ U = \left(\frac 1y\partial_y+1\right)u $$

we have

$$ \left(\frac 1x\partial_x+1\right)U = 0 $$

and solving for $U$ we have

$$ U(x,y) = e^{-\frac{x^2}{2}}\phi(y) $$

etc.

Answer 3

Another method (separation of variables). This is not the most elegant method. Just to not forget it.

One look for particular solutions on the form $\quad u=F(x)G(y)$.

$u_x=F'G\quad;\quad u_y=FG'\quad;\quad u_{xy}=F'G'$. $$F'G'+xyFG+xf'G+yFG'=0$$ $$\left(\frac{F'}{xF}\right)\left(\frac{G'}{yG}\right)+1+\left(\frac{F'}{xF}\right)+\left(\frac{G'}{yG}\right)=0$$ This is possible only if $\left(\frac{F'}{xF}\right)=c_1$ and $\left(\frac{G'}{yG}\right)=c_2$ with $c_1c_2+1+c_1+c_2=0=(c_1+1)(c_2+1)=0.$

First particular solution with $c_1=-1$ :

$\left(\frac{F'}{xF}\right)=-1\quad\implies\quad X(x)=e^{-x^2/2}$ and $G(y)$ any function.

Second particular solution with $c_2=-1$ :

$\left(\frac{G'}{yG}\right)=-1\quad\implies\quad Y(y)=e^{-y^2/2}$ and $F(x)$ any function.

The solution of the ODE is any linear combination of the two particular solutions : $$u(x,y)=e^{-x^2/2}G(y)+e^{-y^2/2}F(x)$$