# Differential Equation $\ln(y') = x - y - e^y$

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 homogeneous. I first tried finding a solution to the associated homogeneous 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-homogeneous equation. Can anyone share a hint or general strategy for this?

Thanks.

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.

• 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? – rehband Jul 8 '14 at 14:49

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 $x=e^y$