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Find an explicit expression for the solution of the IVP

$$ \begin{cases} u_{t}(x,t)+u_{x}(x,t)+u(x,t)=e^{t+2x}\\ \\ u(0,x)=0, \end{cases} $$

by using the method of characteristics

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closed as off-topic by max_zorn, Riccardo.Alestra, mrtaurho, metamorphy, José Carlos Santos Jan 21 at 12:12

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  • $\begingroup$ Hi and welcome to the Math.SE. Questions sounding like "do this for me, please" are poorly received from other members. I you need help, help other users help you for example by explaining what you tried and where did you get stuck. $\endgroup$ – Daniele Tampieri Jan 20 at 17:08
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This is a linear PDE so

$$ u = u^h + u^p $$

with

$$ u_t^h+u_x^h=-u^h\\ u_t^p+u_x^p+u^p=e^{t+2x} $$

Using the Characteristics method with $u^h$ we have

$$ \begin{cases} \frac{dt}{d\tau} = 1 & t(0) = 0 & t(\tau) = \tau\\ \frac{dx}{d\tau} = 1 & x(0) = s & x(\tau) = s+\tau\\ \frac{du^h}{d\tau} = -u^h & u^h(0) = \phi(s) & -\ln u^h(\tau) = \phi(s)+\tau \end{cases} $$

or

$$ u^h(t,x) = e^{-t}\psi(x-t) $$

regarding the particular we have that $u^p(t,x) = \frac 14e^{t+2x}$ verifies the particular equation then

$$ u(t,x) = e^{-t}\psi(x-t)+\frac 14e^{t+2x} $$

and for $t=0$ we have

$$ \psi(x) + \frac 14e^{2x}=0 $$

then $\psi(x) = -\frac 14 e^{2x}$ and finally

$$ u(t,x) = \frac 14\left(-e^{-t}e^{2(x-t)}+e^{t+2x}\right) $$

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Yes, by using the method of characteristics, one can find the explicit expression for the solution : $$u(t,x)=\frac14\left(e^{t+2x}-e^{-3t+2x} \right)$$ The question is concise. The answer as well.

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