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The equation is given as:

$$e^{t^2}u_t+tu_x=0$$ with $u(x,0)=x+2$

I've got $x=\frac{1}{2}e^{t^2}+x_0$ but I'm not sure where to go from there

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I added $\LaTeX$ to your question. Please verify it's correct. –  Pragabhava Feb 12 '13 at 3:47
    
yes this is correct –  user61933 Feb 12 '13 at 4:08
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1 Answer

As you have done, the best way to solve the equation is to write it as $$ t e^{-t^2} u_x + u_t = 0 $$

Let $u = u\big(x(\eta),t(\eta)\big)$, then

$$ \frac{d}{d \eta} u\big(x(\eta),t(\eta)\big) = u_x x' + u_t t' = 0 $$

means that

$$ t'(\eta) = 1, \qquad x'(\eta) = t e^{-t^2}, \qquad u'(\eta) = 0 $$

EDIT

The initial condition is given in the curve $t = 0$, this means we have $t(\eta=0) = 0$.

Now,

$$ t'(\eta) = 1, \quad t(0) = 0 \,\Longrightarrow \,t(\eta) = \eta. $$

Using this, we have for $x$

$$ \frac{d x}{d \eta} = \eta e^{-\eta^2}, $$ and then $$ \int_0^\eta \frac{d x}{d \xi} d\xi = \int_0^\eta \xi e^{-\xi^2} d\xi \quad \Longrightarrow \quad x(\eta) - x(\eta = 0) = \frac{1}{2} \left(1 - e^{-\eta^2}\right) $$

or

$$ x(\eta = 0) = x - \frac{1}{2} \left(1 - e^{-t^2}\right). $$

Finally

$$ \frac{d u}{d \eta} = 0 \quad \Longrightarrow \quad u(\eta) - u(\eta = 0) = 0 $$

or

$$ u(\eta) = u(\eta = 0) = x(\eta = 0) + 2 $$

We now conclude that

$$ u(x,t) = x + 2 - \frac{1}{2} \left(1 - e^{-t^2}\right). $$

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Im a bit confused on how you got the x term. I thought the integral of the te^(-t^2) was just e^-t^2/2. where did the 1 come from? –  user61933 Feb 12 '13 at 4:37
    
also when do we incorporate the fact that when t is 0 x is x+2? –  user61933 Feb 12 '13 at 6:10
    
@user61933 I've edited the answer in order to address your questions. Please, if you find it appropriate, vote up and accept the answer. –  Pragabhava Feb 12 '13 at 18:52
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