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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!

$$u(x,y) = frac{(y+frac{4}{3}x)^3}{12}+sin(y+frac{4}{3}x) - frac{y^3}{12}quadtext{is correct}$$
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.