Finding general solution to this PDE

Question

I need to find all functions $ z = z(y) $ and real constants $ a $ so that $ u(x,y) = z(x+2ay) $ is a solution of the following:

$$ u_{y} = u_{xx} + 4u $$

So I found that $ z = e^{y} $ satisfies the above PDE but how do I find all the others?

Answer 1

If $\ u(x,y)=z(x+ay)\ $, then \begin{align} u_y(x,y)&=az'(z+ay)\ \ \text{ and}\\ u_{xx}(x,y)&=z''(x+ay)\ . \end{align} Therefore $\ u\ $ will satisfy the partial differential equation $\ u_y=u_{xx} + 4u\ $ if and only if $\ z\ $ satisfies the ordinary differential equation $$ z''-az'+4z=0\ . $$ This differential equation has non-trivial real solutions for any value of $\ a\ $, which can be expressed in terms of the roots, $\ r\ $, of its charcteristic equation, $\ r^2-ar+4=0\ $. These roots are $\ r_1= \frac{a+\sqrt{a^2-16}}{2}\ $ and $\ r_2= \frac{a-\sqrt{a^2-16}}{2}\ $, and the form of the real solutions depends on the nature of them.

• If both roots are the same (i.e. $\ a=\pm4\ $) then the general solution is $$ z(y)=(A+By)e^\frac{ay}{2}\ . $$
• If both roots are real, but unequal, (i.e. $\ a^2>16\ $), then the general solution is $$ z(y)=Ae^{\left(\frac{a+\sqrt{a^2-16}}{2}\right)y}+B e^{\left(\frac{a-\sqrt{a^2-16}}{2}\right)y}\ . $$
• If the roots are unequal complex conjugates of each other (i.e. $\ a^2<16\ $) then the general solution is $$ z(y)= e^\frac{ay}{2}\left(A\cos\left(\frac{y\sqrt{16-a^2}}{2}\right)+B \sin\left(\frac{y\sqrt{16-a^2}}{2}\right)\right)\ . $$

Where, in every case, $\ A\ $ and $\ B\ $ are arbitrary real constants. All of the above functions $\ z\ $ will provide you with a solution of the partial differential equation when you put $\ u(x,y)=z(x+ay)\ $, and they are the only ones that will do so.