#### 2.75   ODE No. 75

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

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

$\left \{ y \left ( x \right ) =-{{\rm e}^{x}}+\ln \left ( -1+{{\rm e}^{{{\rm e}^{x}}+{\it \_C1}}} \right ) -{\it \_C1} \right \}$

Hand solution

\begin {align} y^{\prime } & =e^{x-y}-e^{x}\nonumber \\ y^{\prime } & =e^{x}\left ( e^{-y}-1\right ) \nonumber \\ \frac {1}{e^{-y}-1}dy & =e^{x}dx\tag {1} \end {align}

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$\int \frac {1}{u-1}\left ( -\frac {du}{u}\right ) =-\int \frac {1}{u\left ( u-1\right ) }du$ But $$\frac {1}{u\left ( u-1\right ) }=-\left ( \frac {1}{u}-\frac {1}{u-1}\right )$$, hence\begin {align*} -\int \frac {1}{u\left ( u-1\right ) }du & =\int \left ( \frac {1}{u}-\frac {1}{u-1}\right ) du\\ & =\ln u-\ln \left ( u-1\right ) \\ & =\ln e^{-y}-\ln \left ( e^{-y}-1\right ) \\ & =-\left ( \ln \left ( e^{-y}-1\right ) -\ln e^{-y}\right ) \end {align*}

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

Back to (1), when we integrate both sides, and since $$\int e^{x}dx=e^{x}+C$$\begin {align*} -\ln \left ( 1-e^{y}\right ) & =e^{x}+C\\ \ln \left ( 1-e^{y}\right ) & =-e^{x}+C_{1} \end {align*}

Hence\begin {align*} 1-e^{y} & =\exp \left ( -e^{x}+C_{1}\right ) \\ e^{y} & =1-\exp \left ( -e^{x}+C_{1}\right ) \end {align*}

Taking logs$y=\ln \left ( 1-\exp \left ( -e^{x}+C_{1}\right ) \right )$ Let $$e^{C_{1}}=C_{2}$$ then$y=\ln \left ( 1-C_{2}e^{-e^{x}}\right )$ 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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