2.75   ODE No. 75

1. Problem in Latex
2. Mathematica input
3. Maple input

$$y^{\prime }(x)-e^{x-y(x)}+e^x=0$$ ✓ Mathematica : cpu = 0.137958 (sec), leaf count = 20

$$\left\{\{y(x)\to \log (1-e^{-e^x+c_1})\}\right\}$$ ✓ Maple : cpu = 0.116 (sec), leaf count = 20

$$\{y\left(x\right)=-{\mathrm{e}}^x+\ln (-1+{\mathrm{e}}^{{\mathrm{e}}^x+\mathit{\_}\mathit{C}\mathit{1}})-\mathit{\_}\mathit{C}\mathit{1}\}$$

Hand solution

$$\begin{array}{rl}{y}^{\prime }& =e^{x-y}-e^x\\ y^{\prime }& =e^x(e^{-y}-1)\\ \text{(1)}& \frac{1}{e^{-y}-1}dy& =e^{x}dx\end{array}$$

Integrating both sides. $\int \frac{1}{e^{-y}-1}dy$. Let $e^{-y}=u$, then $\frac{du}{dy}=-e^{-y}=-u$. Hence $dy=-\frac{du}{u}$, therefore the integral becomes$$\int \frac{1}{u-1}(-\frac{du}{u})=-\int \frac{1}{u(u-1)}du$$ But $\frac{1}{u(u-1)}=-(\frac{1}{u}-\frac{1}{u-1})$, hence$$\begin{array}{rl}-\int \frac{1}{u(u-1)}du& =\int (\frac{1}{u}-\frac{1}{u-1})du\\ & =\ln u-\ln (u-1)\\ & =\ln e^{-y}-\ln (e^{-y}-1)\\ & =-(\ln (e^{-y}-1)-\ln e^{-y})\end{array}$$

But $\ln x-\ln y=\ln \left(\frac{x}{y}\right)$ and the above becomes$$\begin{array}{rl}\int \frac{1}{e^{-y}-1}dy& =-[\ln \left(\frac{e^{-y}-1}{e^{-y}}\right)]\\ & =-\ln (1-e^y)\end{array}$$

Back to (1), when we integrate both sides, and since $\int e^{x}dx=e^x+C$$$\begin{array}{rl}-\ln (1-e^y)& =e^x+C\\ \ln (1-e^y)& =-e^x+C_1\end{array}$$

Hence$$\begin{array}{rl}1-e^y& =\exp (-e^x+C_1)\\ e^y& =1-\exp (-e^x+C_1)\end{array}$$

Taking logs$$y=\ln (1-\exp (-e^x+C_1))$$ Let $e^{C_1}=C_2$ then$$y=\ln (1-C_2{e}^{-e^x})$$ Verification

ode:=diff(y(x),x)=exp(x-y(x))-exp(x); 
my_sol:=log(1-_C1*exp(-exp(x))); 
odetest(y(x)=my_sol,ode); 
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