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Is there a way to solve this without using laplace transform?: $$\frac{\partial y}{\partial x}= \frac{\partial^2 y}{\partial z^2 }-2ik^2 y$$

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Just to make sure that there are no typos: $y$ is a function of $x$ and $z$, right? –  Dennis Gulko Nov 8 '12 at 12:39
    
yes,thats correct. –  sani Nov 8 '12 at 12:43
    
$x$ and $z$ are real variables? $i=\sqrt{-1}$? What is $k$? –  vesszabo Nov 8 '12 at 17:13
    
well,x and z are real and k is a parameter. solution should be in terms of parameter –  sani Nov 9 '12 at 9:16
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1 Answer

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Unfortunately, this type of PDE is impossible to have non-kernel form solution. So it is unavoidable to solve this type of PDE only by kernel transform, kernel method or separation of variables:

Let $y(x,z)=X(x)Z(z)$ ,

Then $X'(x)Z(z)=X(x)Z''(z)-2ik^2X(x)Z(z)$

$X'(x)Z(z)=X(x)(Z''(z)-2ik^2Z(z))$

$\dfrac{X'(x)}{X(x)}=\dfrac{Z''(z)-2ik^2Z(z)}{Z(z)}=-(f(t))^2-2ik^2$

$\begin{cases}\dfrac{X'(x)}{X(x)}=-(f(t))^2-2ik^2\\Z''(z)+(f(t))^2Z(z)=0\end{cases}$

$\begin{cases}X(x)=c_3(t)e^{-x((f(t))^2+2ik^2)}\\Z(z)=\begin{cases}c_1(t)\sin(zf(t))+c_2(t)\cos(zf(t))&\text{when}~f(t)\neq0\\c_1z+c_2&\text{when}~f(t)=0\end{cases}\end{cases}$

$\therefore y(x,z)=C_1ze^{-2ik^2x}+C_2e^{-2ik^2x}+\int_tC_3(t)e^{-x((f(t))^2+2ik^2)}\sin(zf(t))~dt+\int_tC_4(t)e^{-x((f(t))^2+2ik^2)}\cos(zf(t))~dt~\text{or}~C_1ze^{-2ik^2x}+C_2e^{-2ik^2x}+\sum\limits_tC_3(t)e^{-x((f(t))^2+2ik^2)}\sin(zf(t))+\sum\limits_tC_4(t)e^{-x((f(t))^2+2ik^2)}\cos(zf(t))$

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