Solution to 2nd order PDE What is the general solution to the differential equation:
$$\frac{\partial^2 u}{\partial y^2}=\frac{\partial u}{\partial x}$$
I'm a little stuck because all the techniques I know are unable to solve it.
 A: We can still find its general solution by using separation of variables:
Case 1: $\text{Re}(x)\ge0$
Let $u(x,y)=X(x)Y(y)$ ,
Then $X(x)Y''(y)=X'(x)Y(y)$
$\dfrac{X'(x)}{X(x)}=\dfrac{Y''(y)}{Y(y)}=-(f(t))^2$
$\begin{cases}\dfrac{X'(x)}{X(x)}=-(f(t))^2\\Y''(y)+(f(t))^2Y(y)=0\end{cases}$
$\begin{cases}X(x)=c_3(t)e^{-x(f(t))^2}\\Y(y)=\begin{cases}c_1(t)\sin(yf(t))+c_2(t)\cos(yf(t))&\text{when}~t\neq0\\c_1y+c_2&\text{when}~t=0\end{cases}\end{cases}$
$\therefore u(x,y)=C_1y+C_2+\int_tC_3(t)e^{-x(f(t))^2}\sin(yf(t))~dt+\int_tC_4(t)e^{-x(f(t))^2}\cos(yf(t))~dt$ or $C_1y+C_2+\sum_tC_3(t)e^{-x(f(t))^2}\sin(yf(t))+\sum_tC_4(t)e^{-x(f(t))^2}\cos(yf(t))$
Case 2: $\text{Re}(x)\le0$
Let $u(x,y)=X(x)Y(y)$ ,
Then $X(x)Y''(y)=X'(x)Y(y)$
$\dfrac{X'(x)}{X(x)}=\dfrac{Y''(y)}{Y(y)}=(f(t))^2$
$\begin{cases}\dfrac{X'(x)}{X(x)}=(f(t))^2\\Y''(y)-(f(t))^2Y(y)=0\end{cases}$
$\begin{cases}X(x)=c_3(t)e^{x(f(t))^2}\\Y(y)=\begin{cases}c_1(t)\sinh(yf(t))+c_2(t)\cosh(yf(t))&\text{when}~t\neq0\\c_1y+c_2&\text{when}~t=0\end{cases}\end{cases}$
$\therefore u(x,y)=C_1y+C_2+\int_tC_3(t)e^{x(f(t))^2}\sinh(yf(t))~dt+\int_tC_4(t)e^{x(f(t))^2}\cosh(yf(t))~dt$ or $C_1y+C_2+\sum_tC_3(t)e^{x(f(t))^2}\sinh(yf(t))+\sum_tC_4(t)e^{x(f(t))^2}\cosh(yf(t))$
Hence $u(x,y)=\begin{cases}C_1y+C_2+\int_tC_3(t)e^{-x(f(t))^2}\sin(yf(t))~dt+\int_tC_4(t)e^{-x(f(t))^2}\cos(yf(t))~dt&\text{when}~\text{Re}(x)\geq0\\C_1y+C_2+\int_tC_3(t)e^{x(f(t))^2}\sinh(yf(t))~dt+\int_tC_4(t)e^{x(f(t))^2}\cosh(yf(t))~dt&\text{when}~\text{Re}(x)\leq0\end{cases}$ or $\begin{cases}C_1y+C_2+\sum_tC_3(t)e^{-x(f(t))^2}\sin(yf(t))+\sum_tC_4(t)e^{-x(f(t))^2}\cos(yf(t))&\text{when}~\text{Re}(x)\geq0\\C_1y+C_2+\sum_tC_3(t)e^{x(f(t))^2}\sinh(yf(t))+\sum_tC_4(t)e^{x(f(t))^2}\cosh(yf(t))&\text{when}~\text{Re}(x)\leq0\end{cases}$
This is already the general solution of $\dfrac{\partial^2u}{\partial y^2}=\dfrac{\partial u}{\partial x}$ . Note that when without any I.C.s, the form of $f(t)$ can choose arbitrary, but when I.C.s are given, the form of $f(t)$ and the choice whether using the integration kernel or using the summation kernel should choose wisely in order to accommodate the I.C.s to get the most nice form of the solution, especially the number of I.C.s is more than two.
Another brilliant method is called the power series method.
Similar to PDE - solution with power series:
Let $u(x,y)=\sum\limits_{n=0}^\infty\dfrac{(y-a)^n}{n!}\dfrac{\partial^nu(x,a)}{\partial y^n}$ ,
Then $u(x,y)=\sum\limits_{n=0}^\infty\dfrac{(y-a)^{2n}}{(2n)!}\dfrac{\partial^{2n}u(x,a)}{\partial y^{2n}}+\sum\limits_{n=0}^\infty\dfrac{(y-a)^{2n+1}}{(2n+1)!}\dfrac{\partial^{2n+1}u(x,a)}{\partial y^{2n+1}}=\sum\limits_{n=0}^\infty\dfrac{(y-a)^{2n}}{(2n)!}\dfrac{\partial^nu(x,a)}{\partial x^n}+\sum\limits_{n=0}^\infty\dfrac{(y-a)^{2n+1}}{(2n+1)!}\dfrac{\partial^{n+1}u(x,a)}{\partial x^n\partial y}=\sum\limits_{n=0}^\infty\dfrac{f^{(n)}(x)(y-a)^{2n}}{(2n)!}+\sum\limits_{n=0}^\infty\dfrac{g^{(n)}(x)(y-a)^{2n+1}}{(2n+1)!}$
A: This is the heat equation. Don't expect a single formula for the general solution, because the problem is too complicated for that. There are, however, fairly general formulas which apply under special circumstances (Fourier series solution for the initial–boundary value problem, etc.). 
A: Also you can try self-similar solutions, see for example,
http://www.math.toronto.edu/courses/apm346h1/20129/L9.html
