I am asked to find the general solution $f(x, y)$ of the partial differential equation:

$\frac{\partial ^2 f}{\partial x \partial y}=e ^ {x+2y}$

I know these are relatively easy to solve, I haven't done them in a while and have forgotten how to go about solving them, I haven't yet found an good internet source that explains them straightforwardly.

To attempt a solution, I first found the integral,

$\int e^x e^y dx=e^x e^y +g(y)$

Next, integrating this with respect to $y$,

$\int (e^x e^y +g(y)) \space dy$

solving this becomes,

$ = e^x e^y +yg(y) +h(x)$

Is my reasoning correct? If I integrate a partial derivative with respect to $x$, will the constant become $g(y)$ and if I integrate a partial derivative with respect to $y$, will the content become $h(x)$?

  • 2
    $\begingroup$ More or less it is correct. You forgot a $2$ in the exponent of $e^{2x}$. Also, when you integrate $g(y)$ with respect to $y$ you do not get $yg(y)$ but another function of $y$. $\endgroup$ Aug 25, 2015 at 0:50

1 Answer 1


The methodology in the posted question was correct and gives a way forward. Its implementation had some mistakes which we resolve here.

We begin with

$$\frac{\partial^2 f(x,y)}{\partial x\partial y}=e^{x+2y} \tag 1$$

and integrate $(1)$ with respect to $x$ to obtain

$$\frac{\partial f(x,y)}{\partial y}=e^{x+2y}+C_1(y) \tag 2$$

where $C_1(y)$ is an integration constant.

Next, we integrate $(2)$ with respect to $y$ and obtain

$$f(x,y)=\frac12 e^{x+2y}+\int C_1(y)\,dy+C_2(x)$$

where $C_2(x)$ is a second integration constant.

Finally, labeling $g(y)=\int C_1(y)\,dy$ and $h(x)=C_2(x)$ yields the general result

$$\bbox[5px,border:2px solid #C0A000]{f(x,y)=\frac12 e^{x+2y}+g(y)+h(x)}$$

  • 1
    $\begingroup$ Sorry, fixed one of your equations to see it, had to add other dollar signs for them to accept the edit, might make it more confusing for other viewers to understand, but I understand the solution now, thank you. $\endgroup$
    – mnmakrets
    Aug 25, 2015 at 4:05
  • 1
    $\begingroup$ You're more than welcome! My Pleasure. $\endgroup$
    – Mark Viola
    Aug 25, 2015 at 4:47

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