I need to solve this linear PDE:
$3u_x - 4u_y = y^2$
The initial condition provided is:
$ u (0,y)= sin(y)$
I need to use the Characteristic Method. I learned the method from this video.
I have reached an answer. However, I am not sure if it is wright.
My intermediate steps are:
First constant: $c_1= y + \frac{4}{3}x $
Second constant: $c_2= \frac{y^3}{3} + 4u $
Using an arbitrary function G to make the relation between both constants,
$c_2 =G(c_1) $, we have that:
$\frac{y^3}{3} + 4u = G(y + \frac{4}{3}x) $
With the initial condition we have:
$G(y) = \frac{y^3}{3} +4sin(y)$
After the definition of $G(y)$ above , I inputed the value of $c_1$ , having:
$G(y + \frac{4}{3}x) = \frac{(y+\frac{4}{3}x)^3}{3}+ 4sin(y+\frac{4}{3}x) $.
Finally, solving for $u$:
$u(x,y) = \frac{(y+\frac{4}{3}x)^3}{12}+sin(y+\frac{4}{3}x) - \frac{y^3}{12}$
A friend of mine solved this problem with a different approach. She reached a different result. There are some comments along her solution that were written in portuguese.
Is this right?
If I did something wrong, what was it?
Thanks in advance!
$$u(x,y) = \frac{(y+\frac{4}{3}x)^3}{12}+\sin(y+\frac{4}{3}x) - \frac{y^3}{12}\quad\text{is correct}$$ Expanding leads to : $$u(x,y)=\sin(y+\frac{4}{3}x)+\frac{y^2x}{3}+\frac{4yx^2}{9}+\frac{16x^3}{81}$$ So, there is no mistake in your calculus. There is a sign mistake in the handwritten page, which at end gives $\sin(y+\frac{4}{3}x)-\frac{y^2x}{3}+\frac{4yx^2}{9}-\frac{16x^3}{81}$.
Unfortunately the handwritten page is not enough readable to see where exactly the mistake occurred.