A mid-step in solving a pde requires integrating the function $f'(-x) = -2x + e^x$ with respect to $x$. What do you do differently when integrating a function of $-x$ rather than of $x$? Am I allowed to set say $y = -x$ and so it becomes $f'(y) = 2y + e^{-y}$ which gives $f(y) = y^2 - e^{-y} + c$ and then substitute back in $-x$?
Mid-step in solving a pde
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real-analysis
pde
2 Answers
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$f(x) = \int f'(x) \mathrm{d}x = -\int f'(-x)\mathrm{d}x = \int (2x-e^{x})\mathrm{d}x = x^2 - e^{x} + C$
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0@morphism it was a silly error, I corrected it. – 2012-05-19
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I think you missed something in the integral. Integrating with a function of $-x$ instead of $x$ is done by changing either the function's variable $(-x)$ or the differential $(dx)$, so that they match. $\int f'(-x) dx=\int -f'(y)dy$ since $y=-x \Leftrightarrow dy=-dx$. You got:
$\int -f'(y)dy=\int 2y+e^{-y}dy=y^2-e^{-y}+c=x^2-e^x+c$
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1$\int 2y dy = y^2$ , you should correct that – 2012-05-18