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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.

enter image description here enter image description here Is this right?

If I did something wrong, what was it?

Thanks in advance!

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    $\begingroup$ It is right. What was the result she got? $\endgroup$ – Rafa Budría Nov 14 '18 at 19:25
  • $\begingroup$ @RafaBudría, just updated my post. Thanks for checking my result. $\endgroup$ – Pedro Delfino Nov 14 '18 at 19:54
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    $\begingroup$ Except the signs for the terms it is the same expression. The error was not carry the minus sign in $t=-\bar x/3$ along. $\endgroup$ – Rafa Budría Nov 14 '18 at 20:57
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    $\begingroup$ General solution of equation is $u=F(y+\frac{4}{3}x)-\frac{y^2}{12}$. $\endgroup$ – Aleksas Domarkas Nov 15 '18 at 13:34
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$$u(x,y) = \frac{(y+\frac{4}{3}x)^3}{12}+\sin(y+\frac{4}{3}x) - \frac{y^3}{12}\quad\text{is correct}$$ 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.

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