Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
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
up vote 0 down vote accepted

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





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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.