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Find the solution to this initial value problem on the largest interval. $$ln(y') = x - y - e^y, \,\,\,\,\,\,\,\,\,\,\,\,y(1)=0.$$

So this differential equation is not linear and not homogenous. I first tried finding a solution to the associated homogenous equation $$ln(y') = - y - e^y$$$$\iff y' =e^{-(y+e^y)}$$ which I was able to solve by separating the variables. The general solution I thus found is $$y(x) = C \,\, ln(ln(x)).$$

Now I wonder how to find the solution to the original non-homogenous equation. Can anyone share a hint or general strategy for this?

Thanks.

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First thing to notice is $y = \ln\left(\mathrm{e}^{y}\right)$ $$ \ln(y') = x - \ln\left(\mathrm{e}^{y}\right) -\mathrm{e}^{y} $$ then we have $$ \ln\left(y'\mathrm{e}^{y}\right) = x - \mathrm{e}^{y} $$ using the sub $v = \mathrm{e}^{y}$ leads to $$ \ln(v') = x- v $$

thus $$ v' = \mathrm{e}^{x}\mathrm{e}^{-v} $$

hence $$ \mathrm{e}^{v} = \mathrm{e}^{x}+C\implies v = \ln\left(\mathrm{e}^{x}+C\right) $$ and subbing in for y $$ y(x) = \ln\left[\ln\left(\mathrm{e}^{x}+C\right)\right] $$ now we have $y(1) = 0$ which means $$ y(1) = 0 = \ln\left[\ln\left(\mathrm{e}+C\right)\right] $$ therefore $C = 0$ so the solution is actually $$ y(x) = \ln(x) $$ you could check that the solution you found does not hold for the original equation.

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  • $\begingroup$ Ahh thanks, that makes a lot of sense. Now how would I determine the largest interval on which y is a solution of the differential equation? I assume it would be $(0, \infty)$, right? $\endgroup$ – rehband Jul 8 '14 at 14:49
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It seems like the most sensible first step is to remove that logarithm, at which point the equation is separable.

$$y'=\dfrac{e^x}{e^{y+e^y}}$$ $$e^ye^{e^y}dy=e^xdx$$ $$e^{e^y}=e^x+C$$ $$e^{e^0}=e^1+C,C=0$$

So the solution looks to be $x=e^y$

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