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Could someone help me with this question?

Find the general solution of $\;(1+e^x)y\,y' = e^x$

I tried to divide both sides by $\;(1+e^x)\,y,\;$ but I found nothing.

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    $\begingroup$ Try dividing by $(1+e^x)$ instead. It is separable. $\endgroup$ Aug 24 '13 at 23:18
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It is separable. Divide both sides by $1+e^{x}$ to get $$yy'=\frac{e^x}{e^x+1}\Longleftrightarrow \left(\frac{y^2}{2}\right)'=\left(\ln(1+e^x)\right)',$$ hence the general solution is $$\frac{y^2}{2}=\ln(1+e^x)+\mathrm{const}.$$

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The equation is separable. Divide both sides by $(1 + e^x)$.

$$(1+e^x)yy' = e^x\iff y \,dy = \frac {e^x}{1 + e^x}\,dx$$

Now integrate each side of the equation, and you're done. For the right hand side, we have $u = 1 + e^x \implies \,du = e^x \,dx$

$$\begin{align} \int y\,dy & = \int \frac {\overbrace{e^x\,dx}^{du}}{\underbrace{1 + e^x}_{u}} \\ \\ \iff \dfrac {y^2}2 & = \ln (1 + e^x) + C \\ \\ \iff y^2 & = 2 \ln(1+e^x) + C' \end{align}$$

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    $\begingroup$ I find the last step rather unnecessary. $\endgroup$
    – Pedro Tamaroff
    Aug 25 '13 at 0:31
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    $\begingroup$ @amWhy: Nice use of braces for explaining +1 $\endgroup$
    – Amzoti
    Aug 25 '13 at 0:43
  • $\begingroup$ @Peter I agree; I've removed it accordingly. $\endgroup$
    – amWhy
    Aug 25 '13 at 14:35
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We can also change variables in the first place. Let's define $u \equiv e^x$. Now the equation becomes:

$$(1+u)y\frac{dy}{du}\frac{du}{dx}=u$$

But, $\dfrac{du}{dx}=e^x=u$, therefore we will have:

\begin{align} \begin{aligned} (1+u)y\frac{dy}{du}&=1 \\ \Rightarrow y \ dy &=\frac{du}{1+u} \\ \Rightarrow \frac{y^2}{2}&=\ln(1+u) + C \\ &=\ln (1+e^x)+C \end{aligned} \end{align}

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