# Solve this Semi-Linear PDE (Partial Differential Equation) with the Characteristic Method

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?

• 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. – Rafa Budría Nov 14 '18 at 20:57
• General solution of equation is $u=F(y+\frac{4}{3}x)-\frac{y^2}{12}$. – Aleksas Domarkas Nov 15 '18 at 13:34
$$u(x,y) = \frac{(y+\frac{4}{3}x)^3}{12}+\sin(y+\frac{4}{3}x) - \frac{y^3}{12}\quad\text{is correct}$$ 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 $$\sin(y+\frac{4}{3}x)-\frac{y^2x}{3}+\frac{4yx^2}{9}-\frac{16x^3}{81}$$.