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I need help with this calculus problem:

Find the particular solution of the differential equation

$e^y\frac{dy}{dx}=e^{−9x}$, such that $y=7$ when $x=0$

I got $-\ln(\frac{e^{-9x}}{9}-\frac{1}{9}-e^7)$

But its wrong can someone work through the problem and show me what I did wrong ?

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  • $\begingroup$ This equation is separable, try integrating both sides of the following, $e^ydy = e^{-9x}dx$ $\endgroup$ – Tony S.F. Jun 21 '16 at 17:29
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Hint $$e^y\frac{\mathrm d y}{\mathrm d x}=e^{-9x}\implies e^d\mathrm d y=e^{-9x}\mathrm d x\implies \int e^y\mathrm d y=\int e^{-9x}\mathrm d x.$$

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  • $\begingroup$ i did this, and i still got the wrong answer $\endgroup$ – google Jun 21 '16 at 17:36
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$e^y = -\frac 19 e^{-9x} + C\\ e^7 = -\frac 19 + C\\ C = e^7 + \frac 19\\ e^y = -\frac 19 e^{-9x} + e^7 + \frac 19$

And, I think you had something close to this.

$y = \ln (-\frac 19 e^{-9x} + e^7 + \frac 19)$

Now, how do what to simplify that? $y = \ln (-e^{-9x} + e^7 + 1) - \ln 9$

There are other operations we an perform, but I don't think it gets much simpler than that.

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  • $\begingroup$ but its still saying this solution is wrong. i mean i dont know what else to try, you got the same answer i did just more simplified $\endgroup$ – google Jun 21 '16 at 17:54
  • $\begingroup$ @google What is saying this is wrong? You can plug this into wolfram alpha and it gets the same answer as all three of us. $\endgroup$ – qbert Jun 21 '16 at 18:04
  • $\begingroup$ Your answer has a negative sign in front of the logarithm. Mine has it inside. $\endgroup$ – Doug M Jun 21 '16 at 18:28
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Fortunately, this is a separable equation. $$\frac{dy}{dx}e^y=e^{-9x}\Rightarrow e^ydy=e^{-9x}dx\Rightarrow \int e^{y}dy=\int e^{-9x}dx\Rightarrow e^y=\frac{-e^{-9x}}{9}+c$$ Then, with the boundary condition $y(0)=7$ we get $$e^7=\frac{-1}{9}+c\Rightarrow c=e^7+1/9 $$ Yielding the IVP solution: $$e^y=\frac{e^{-9x}}{-9}+e^7+\frac{1}{9}\Rightarrow y(x)=\ln(e^7+\frac{1-e^{-9x}}{9})$$

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$e^y \frac{dy}{dx} = e^{-9x}\Rightarrow e^y = -\frac{1}{9} e^{-9x} + C \Rightarrow y = \ln (-\frac{1}{9}e^{-9x}+C)=C (\ln 9-9x) \Rightarrow C = \frac{7}{\ln 9} $

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  • $\begingroup$ Not the same 'C's but both constant anyway $\endgroup$ – Ronny Landsverk Jun 21 '16 at 18:21

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