Question

In Problems 3–6 the dependent variable y is missing in the given differential equation. Proceed as in Example 1 and solve the equation by using the substitution u = y'.

6.

${\mathit{e}}^{\mathbf{-}\mathbf{x}}{\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{=}{\left(y^{\text{'}}\right)}^{\mathbf{2}}$

Short answer

The solution is

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

Step-by-step solution

given information

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

solve the equation

We have the second order differential equation

$e^{-x}{y}^{\text{'}\text{'}}={\left(y^{\text{'}}\right)}^2....\left(1\right)$

and we have to solve it as the following technique:

First, we have the substitution

$y^{\text{'}}=u.....\left(2\right)$

After that, since we have y' = u and y is a dependent variable then by differentiation with respect to we can have

$y^{\text{'}\text{'}}=u^{\text{'}}.....\left(3\right)$

Second, substitute from equations (2) and (3) with (y') and (y") into the given equation (1) , then we obtain

role="math" localid="1668066162077" $$\begin{gathered} e^{-x}{u}^{\text{'}}=(u{)}^2 \\ {e}^{-x}\frac{du}{dx}=u^2 \\ \frac{du}{u^2}=e^{x}dx \end{gathered}$$

After that, we have to separate variables and do integration as

$$\begin{gathered} \int \frac{du}{u^2}=\int e^{x}dx \\ -\frac{1}{u}=e^x+c \\ u=-\left(\frac{1}{e^x+c}\right).....\left(4\right) \end{gathered}$$

substitution

Third, substitute from equation (2) with into equation (4), then we have

$$\begin{gathered} y^{\text{'}}=-\left(\frac{1}{e^x+c}\right) \\ \frac{dy}{dx}=-\left(\frac{1}{e^x+c}\right) \end{gathered}$$

After that, we have to separate variables and do integration as

$$\begin{gathered} dy=-\left(\frac{1}{e^x+c}\right)dx \\ \int dy=-\int \left(\frac{1}{e^x+c}\right)dx \\ y=-\int \left(\frac{1}{e^x+c}\right)dx.....\left(5\right) \end{gathered}$$

This integration $\int \left(\frac{1}{e^x+c}\right)dx$can be made using substitution as the following

If we put $s=e^x+c$, then we have

$ds=e^{x}dx,\left(dx=e^{-x}ds=\frac{1}{s-c}ds\right)$

Then we obtain

$$\begin{gathered} \int \left(\frac{1}{e^x+c}\right)dx=\int \frac{1}{s}\frac{1}{s-c}ds \\ =\int \frac{1}{s(s-c)}ds \end{gathered}$$

Now we have to do partial fractions to this fraction as the following

$$\begin{gathered} \frac{1}{s(s-c)}=\frac{A}{s}+\frac{B}{s-c} \\ =\frac{A(s-c)+Bs}{s(s-c)} \\ =\frac{As-Ac+Bs}{s(s-c)} \\ =\frac{(A+B)s-Ac}{s(s-c)} \end{gathered}$$

substitute constants

Then we have

$$\begin{gathered} -Ac=1then:A=-\frac{1}{c} \\ A+B=0then:B=\frac{1}{c} \end{gathered}$$

Substitute with the constants A and B in the fractions, then we have

$\frac{1}{s(s-c)}=-\frac{1}{c}\frac{1}{s}+\frac{1}{c}\frac{1}{s-c}$

Then we obtain the required integration as

$$\begin{gathered} \int \frac{1}{s(s-c)}du=\int \left(-\frac{1}{c}\frac{1}{s}+\frac{1}{c}\frac{1}{s-c}\right)ds \\ =-\frac{1}{c}\int \frac{1}{s}ds+\frac{1}{c}\int \frac{1}{s-c}ds \\ =-\frac{1}{c}lns+\frac{1}{c}ln(s-c)+lnk \end{gathered}$$

Since we have $s=e^x+c$, then we have

$\int \left(\frac{1}{e^x+c}\right)dx=-\frac{1}{c}ln\left(e^x+c\right)+\frac{1}{c}ln\left(e^x\right)+lnk$

After that, substitute with the result of this integration into equation ( 5), then we obtain

$$\begin{gathered} =\frac{1}{c}ln\left(e^x+c\right)-\frac{1}{c}ln\left(e^x\right)-lnk \\ =\frac{1}{c}\left[ln\left(e^x+c\right)-ln\left(e^x\right)\right]-lnk \\ =\frac{1}{c}ln\left(\frac{e^x+c}{e^x}\right)-lnk \\ =\frac{1}{c}ln\left(1+c{e}^{-x}\right)-lnk \end{gathered}$$

conclusion

The solution is

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