Prompt: y'=e^(x-y); y(0)=1 I understand what to do for the particular solution but am having trouble finding the general equation.
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Originally Posted by Lamont Prompt: y'=e^(x-y); y(0)=1 I understand what to do for the particular solution but am having trouble finding the general equation. The equation is seperable! $\displaystyle \frac{dy}{dx}=e^{x}e^{-y} \iff e^{y}dy=e^{x}dx \iff e^{y}=e^{x}+C$ Now just plug in your point to get $\displaystyle e=e^{0}+C \implies C=e-1$ This gives $\displaystyle e^y=e^x+e-1 \implies y=\ln(e^x+e-1)$
Thanks!
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