solve the differential equation $e^y(y'+1)=e^x$

Question

$e^y(y'+1)=e^x$

solve the differential equation

Answer 1

As mentioned in the comments, you can use the substitution $u=\mathrm{e}^y$, and then you'll get $y'\mathrm{e}^y=u'$. The equation now becomes $$u'+u=\mathrm{e}^x$$ which its solution is $$u=C\mathrm{e}^{-x}+\frac12\mathrm{e}^x$$ Finally $y$ would be $$y=\log{u}=\log{(C\mathrm{e}^{-x}+\frac12\mathrm{e}^x)}$$

Answer 2

Another method called reduction to separation of variables is to spot when a differential equation is of the form $y'=f(Ax+By+c)$. Then you substitute $u=Ax+By+c$.

Here we have $$y'=e^{x-y}-1$$

Trying $u=x-y$, gives $u'=1-y'$, or $y'=1-u'$ so we have:

$$1-u'=e^u-1$$ or:

$$u'=2-e^u \implies \int\frac{du}{2-e^u}=x+c$$

Make the sub $v=2-e^u$, so $dv=-e^u du$ or $-e^{-u}dv=du$ or $$\frac{dv}{v-2}=du$$

Then $$\int \frac{dv}{v(v-2)}=\frac{1}{2}\int \frac{1}{v-2}-\frac{1}{v} dv=x+c$$

Or $$\frac{1}{2}(\ln(v-2)-\ln(v))=x+c$$

$$\frac{1}{2}\ln\left(\frac{e^u}{e^u-2}\right)=x+c$$

$$\frac{e^{x-y}}{e^{x-y}-2}=Ae^{2x}$$

$$\frac{e^{x}}{e^{x}-2e^y}=Ae^{2x}$$

$$e^{-x}=A\left(e^x-2e^{y}\right)$$

$$\frac{1}{2}\left(e^x-Be^{-x}\right)=e^y$$

Or to be consistent with the other solution:

$$y=\ln\left(\frac{1}{2}e^x+Ce^{-x}\right)$$