General solution for differential equation $px-qy=2xe^{-(x^2+y^2)}$

Question

Show that general solution of $$px-qy=2xe^{-(x^2+y^2)}$$ can be expressed in the form of $$z=e^{2xy}\int_0^{x+y}e^{-u^2}du+e^{-2xy}\int_0^{x+y}e^{-u^2}du+f(x,y)$$

Where $p=\frac{\partial z}{\partial x}, q=\frac{\partial z}{\partial y}$

From Lagrangian equations:

$$\frac {dx}{x}=\frac {dy}{-y}=\frac {dz}{2x e^{-(x^2+y^2)}}$$

Taking first two terms we obtain $$xy=c_1$$ Then taking second and third terms, and putting $x= \frac{c_1}{y}$ we get

$$dz+\frac{2c_1 e^{-\left(y^2+\frac{c_1^2}{y^2}\right)}}{y^2} dy= 0$$

How do I proceed from here?

Answer 1

Hint.

Making the variables change

$$ \xi = x y\\ \eta = \frac xy $$

we obtain

$$ \eta u_{\eta}(\xi ,\eta )+\sqrt{\eta \xi } e^{-\frac{\left(\eta ^2+1\right) \xi }{\eta }}=0 $$

and then

$$ u(\xi,\eta) = \frac{1}{2} \sqrt{\pi } e^{-2 \xi } \left(\text{erf}\left(\left(\frac{1}{\eta }-1\right) \sqrt{\eta } \sqrt{\xi }\right)-e^{4 \xi } \text{erf}\left(\left(\frac{1}{\eta }+1\right) \sqrt{\eta } \sqrt{\xi }\right)+e^{4 \xi }-1\right)+\phi(\eta) $$

or simplifying a little more

$$ u(\xi,\eta) = \frac{1}{2} \sqrt{\pi } e^{-2 \xi } \left(e^{4 \xi } \text{erfc}\left((\eta +1) \sqrt{\frac{\xi }{\eta }}\right)-\text{erfc}\left((1-\eta ) \sqrt{\frac{\xi }{\eta }}\right)\right)+\phi(\xi) $$

NOTE

The final integration was obtained with the MATHEMATICA command

Integrate[-E^(-(((1 + eta^2) xi)/eta)) Sqrt[xi/eta], eta] // FullSimplify

After recovering the original variables, this integral is equivalent to

$$ \int\frac{e^{-\frac{C_1^2}{x^2}-x^2}}{x^2}dx $$

now as

$$ \left(\frac{C_1}{x}+x\right)^2-2C_1 = \frac{C_1^2}{x^2}+x^2 $$

the integral becomes

$$ e^{2C_1}\int\frac{e^{-\left(\frac{C_1}{x}+x\right)^2}}{x^2}dx $$

I hope this helps.

Concerning the above integral we have that

$$ \left(\int_0^x e^{a-\left(\frac{c^2}{x}+x\right)^2}\right)' = e^{a-\left(\frac{c^2}{x}+x\right)^2} \left(1-\frac{c^2}{x^2}\right)\\ \left(\int_0^x e^{-a-\left(\frac{c^2}{x}-x\right)^2}\right)' = -e^{-a-\left(\frac{c^2}{x}+x\right)^2} \left(1+\frac{c^2}{x^2}\right) $$

and

$$ \left(\int_0^x e^{-a-\left(\frac{c^2}{x}-x\right)^2}\right)' -\left(\int_0^x e^{a-\left(\frac{c^2}{x}+x\right)^2}\right)' = \frac{\left(c^2-x^2\right) e^{a-\left(\frac{c^2}{x}+x\right)^2}}{x^2}+\frac{\left(c^2+x^2\right) e^{-a-\left(x-\frac{c^2}{x}\right)^2}}{x^2} $$

now choosing a convenient $a$ ...