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I have to solve the following pde using Laplace transforms:

$xw_x + w_t= xt$ i.c: w(x,0)= 0

Firstly, transforming the above wrt t, i get: $\bar{w_x} + s\bar{w}/x = 1/s^2$

But, in the textbook, the transformation is given as : $\bar{w_x} + s/x = 1/s^2$ Why is there no $\bar{w}$ on the LHS in the textbook answer?

definition of Laplace transform from textbook: $\int_{0}^{\infty} e^{-st}u(t)dt$

Lalpace transform used: $L(u')= s\bar{u} - u_0$

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  • $\begingroup$ Welcome to Math.SE! It would be helpful if you would include your definition of Laplace transform. Could you do that? $\endgroup$
    – Hrodelbert
    Apr 27, 2015 at 12:59
  • $\begingroup$ That is not a good definition of the Laplace transform, it should be regarded as a theorem which follows from the actual definition. That definition should be $L(u)(s)=\int_0^\infty e^{-st} u(t) dt$. Without that, where would the $1/s^2$ come from? $\endgroup$
    – Ian
    Apr 27, 2015 at 13:09

1 Answer 1

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Solve ODE $$\bar{w}_x + s\bar{w}/x = 1/s^2,\;\bar{w}(0)=0$$ $$\bar{w}=\frac{x}{{{s}^{3}}+{{s}^{2}}}=\frac{x}{s+1}-\frac{x}{s}+\frac{x}{{{s}^{2}}}$$ Answer: $$w={{e}^{-t}} x+t x-x$$

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