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$\frac{\partial^2 z}{\partial x^2}+z=0$, given that at $x=0, z=e^y$ and $\frac{\partial z}{\partial x}=1$

I am doing it in following way:

Integrating w.r.t x twice

$\implies \frac{\partial z}{\partial x} +zx=f(y)$, where f(y) is an arbitrary function ...(1)

$\implies z +\frac {zx^2}{2}=xf(y)+g(y) $, where g(y) is an arbitrary function ...(2)

Putting $x=0, z=e^y$ and $\frac{\partial z}{\partial x}=1$ in (1) and (2)

$\implies e^y=g(y)$ and $f(y)=1$

$\implies z(1+\frac {x^2}{2})=x+e^y $

But the answer to this question is given as $z=sinx+e^ycosx$.

Can someone tell me how to solve this PDE by direct integration only?

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    $\begingroup$ $z$ is supposed to be a function of $x$. The integral of $z$ w.r.t. $x$ is not $zx$. $\endgroup$ Dec 15, 2019 at 5:15

1 Answer 1

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Just pretend that $y$ is a constant and solve the equation as if $z$ is a function of $x$ alone. Then the solution is $A \cos x+B\sin x$ where $A$ and $B$ are constants. Now if you consider $y$ as a variable then you have to replace $A$ and $B$ by functions of $y$. The initial conditions easily give you $A=e^{y}$ and $B=1$ so the solution is $e^{y} \cos x+\sin x$.

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  • $\begingroup$ The general solution of the ODE $y''+y=0$ is $y=A\cos x+B\sin x$. I am assuming that you know this already since this is only an ODE and only one variable calculus is involved. @ShivaneeGupta $\endgroup$ Dec 15, 2019 at 5:31

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