Solving PDE Problem: du/dt+du/dx=0 with Initial Condition u(x,0)=xe-x2

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Homework Statement



which solutions of du/dt+du/dx=0 is equal to xe-x2

Homework Equations


The Attempt at a Solution



u(x,0) = xe-x2
u(x,t)= (x-t)e(-x-t)2

what else do i need to do?
 
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pentazoid said:

Homework Statement



which solutions of du/dt+du/dx=0 is equal to xe-x2

Homework Equations





The Attempt at a Solution



u(x,0) = xe-x2
u(x,t)= (x-t)e(-x-t)2

what else do i need to do?
One way of solving a pde like that is to always try separation of variables , so that you assumeu(x,t)=X(x)T(t)where X is just a function of x and T just a function of t, plug this into your pde and see what u get!
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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