I want to find the solution of $$ \frac{\partial u}{\partial y} = \frac{\partial^2 u}{\partial x^2}$$ I know I can solve it if there were some initial conditions but I want to find the general solution for it, I tried a linear change of variable but it didn't work.

  • 5
    $\begingroup$ This is heat equation. $\endgroup$ – user658409 Aug 31 '19 at 13:46
  • $\begingroup$ Hint: Fundamental solution $\endgroup$ – I was suspended for talking Aug 31 '19 at 13:49

Similar to PDE - solution with power series:

Let $u(x,y)=\sum\limits_{n=0}^\infty\dfrac{(x-a)^n}{n!}\dfrac{\partial^nu(a,y)}{\partial x^n}$ ,

Then $u(x,y)=\sum\limits_{n=0}^\infty\dfrac{(x-a)^{2n}}{(2n)!}\dfrac{\partial^{2n}u(a,y)}{\partial x^{2n}}+\sum\limits_{n=0}^\infty\dfrac{(x-a)^{2n+1}}{(2n+1)!}\dfrac{\partial^{2n+1}u(a,y)}{\partial x^{2n+1}}=\sum\limits_{n=0}^\infty\dfrac{(x-a)^{2n}}{(2n)!}\dfrac{\partial^nu(a,y)}{\partial y^n}+\sum\limits_{n=0}^\infty\dfrac{(x-a)^{2n+1}}{(2n+1)!}\dfrac{\partial^{n+1}(a,y)}{\partial y^n\partial x}=\sum\limits_{n=0}^\infty\dfrac{f^{(n)}(y)(x-a)^{2n}}{(2n)!}+\sum\limits_{n=0}^\infty\dfrac{g^{(n)}(y)(x-a)^{2n+1}}{(2n+1)!}$


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