Answer check transport equation (1st order linear PDE)

[jianxu]

Homework Statement

Hi everyone, I just wanted to double check if I've solved this correctly?

Given:
\left.\frac{du}{dx} + sin(x)\frac{du}{dy} = 0

\left.-\infty < x < \infty

y > 0

\left.u(\frac{\pi}{2} , y ) = y^{2}

Solve the PDE

Homework Equations

Method of characteristics

The Attempt at a Solution

Using method of characteristics, first I said:

\left.\frac{dy}{dx}= \frac{sin x}{1}

taking the integral I get
\left. y = -cos(x) + C

solving for C:
C = y + cos(x)

so now,
\left.u(x,y) = f(y+cos(x))

our initial condition gives us the following:
\left.u(\frac{\pi}{2},y) = y^{2}
so:
\left.u(\frac{\pi}{2},y) = f(y)

which means:
\left. f(y) = y^{2}

Therefore:
\left. u(x,y) = f(y + cos x) = (y+cos(x))^{2}

I think this should be right but I wanted to double check. Thanks!

 

[Mark44]

Why don't you, then? (Double check, that is.) One thing you should get in the habit of doing is checking your own work. Take the partials of u(x, y) and see if they satisfy your differential equation and initial conditions. If they do, you're golden.

 

[jianxu]

hm didn't think about that before. thanks