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

Question

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.

Answer 1

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.

Answer 2

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 $x=e^y$