Question

In Problems 19–30 proceed as in Example 5 to ­find a solution of the given initial-value problem.$\mathbf{26}\mathbf{.}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{3}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathit{y}\mathbf{=}\frac{\mathbf{1}}{\mathbf{1}\mathbf{+}{\mathbf{e}}^{\mathbf{x}}}\mathbf{,}\mathit{y}\left(0\right)\mathbf{=}\mathbf{0}\mathbf{,}\mathit{y}\mathbf{\text{'}}\left(0\right)\mathbf{=}\mathbf{1}$

Short answer

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

Step-by-step solution

The solve differential equation

We have the second order differential equation

$y^{\text{'}\text{'}}+3y^{\text{'}}+2y=\frac{1}{1+e^x}$

with the initial conditions

$y\left(0\right)=0\text{and}{y}^{\text{'}}\left(0\right)=1$

and we have to get its solution as the following technique:

First, we have to find the solutions of the corresponding homogeneous differential equation

$y^{\text{'}\text{'}}+3y^{\text{'}}+2y=0$

by assuming that$y=e^{mx}$is a solution for this homogeneous differential equation, then follow then differentiate this assumption with respect to $x$, then we have

$$\begin{gathered} y^{\text{'}}=m{e}^{mx} \\ {y}^{\text{'}\text{'}}=m^2{e}^{mx} \end{gathered}$$

After that, substitute with$y=e^{mx}$and equation (2) into the differential equation (1), then we obtain

$$\begin{gathered} m^2{e}^{mx}+3m{e}^{mx}+2 \\ {e}^{mx}=0e^{mx}\left(m^2+3m+2\right)=0 \end{gathered}$$

Since$e^{mx}$cannot be equal 0, the we have

$$\begin{gathered} m^2+3m+2=0 \\ (m+1)(m+2)=0 \end{gathered}$$

Then the roots are

$m_1=-1\text{and}{m}_2=-2$

which are real and repeated.

Then we can obtain the solution of the corresponding homogeneous differential equation as

$y_c=k_1{e}^{-x}+k_2{e}^{-2x}..............\left(a\right)$

Second, now we have to find the particular solution $y_p\left(x\right)$of the given differential equation as the following technique:

Since we have $y_1=e^{-x}and{y}_2=e^{-2x}$, then we can obtain the wronskian$W\left(y_1,y_2\right)$ as

$$\begin{gathered} W\left(e^{-x},e^{-2x}\right)=\left|\begin{array}{cc}{y}_1& y_2\\ y_1\text{'}& y_2\text{'}\end{array}\right| \\ =\left|\begin{array}{cc}{e}^{-x}& e^{-2x}\\ -e^{-x}& -2e^{-2x}\end{array}\right| \\ =\left(e^{-x}\right)\times \left(-2e^{-2x}\right)-\left(e^{-2x}\right)\times \left(-e^{-x}\right) \\ =-2e^{-3x}-\left(-e^{-3x}\right) \\ =-2e^{-3x}+e^{-3x} \\ =-e^{-3x} \end{gathered}$$

After that, we have to find the Green's function for the given differential equation as

$$\begin{gathered} G(x,t)=\frac{y_1\left(t\right)y_2\left(x\right)-y_1\left(x\right)y_2\left(t\right)}{W\left(t\right)} \\ =\frac{e^{-t}\times e^{-2x}-e^{-x}\times e^{-2t}}{-e^{-3t}} \\ =\frac{e^{-x}\times e^{-2t}-e^{-t}\times e^{-2x}}{e^{-3t}} \\ =\frac{e^{-x}\times e^{-2t}-e^{-t}\times e^{-2x}}{e^{-3t}}\times \frac{e^{3t}}{e^{3t}} \\ =e^{-x}{e}^t-e^{-2x}{e}^{2t} \end{gathered}$$

Using the Green’s function

Then, since $f\left(t\right)=\frac{1}{1+e^t}$and using the Green's function we have to find the particular solution $y_p\left(x\right)$as

$$\begin{gathered} y_p\left(x\right)={\int }_0^{x}G(x,t)f\left(t\right)dt \\ ={\int }_0^x\left(e^{-x}{e}^t-e^{-2x}{e}^{2t}\right)\frac{1}{1+e^t}dt \\ =e^{-x}{\int }_0^x\frac{e^t}{1+e^t}dt-e^{-2x}{\int }_0^x\frac{e^{2t}}{1+e^t}dt \\ =e^{-x}{\left[\ln \left(1+e^t\right)\right]}_0^x-e^{-2x}{\int }_0^x\frac{e^{2t}}{1+e^t}dt \end{gathered}$$

The integration$\int \frac{e^{2t}}{1+e^t}dt$can be made by substitution as the following:

if we set $\left(u=e^t\right)$, then $\left(du=e^{t}dt\right)$, then we have

$$\begin{gathered} \int \frac{e^{2t}}{1+e^t}dt=\int \frac{e^t{e}^{t}dt}{1+e^t} \\ =\int \frac{u}{1+u}du \\ =\int \frac{1+u-1}{1+u}du \\ =\int du-\int \frac{1}{1+u}du \\ =u-\ln (1+u) \\ =e^t-\ln \left(1+e^t\right) \end{gathered}$$

Then we have the particular solution of the given differential equation as

$$\begin{gathered} y_p\left(x\right)=e^{-x}{\left[\ln \left(1+e^t\right)\right]}_0^x-e^{-2x}{\left[e^t-\ln \left(1+e^t\right)\right]}_0^x \\ =e^{-x}\left[\ln \left(1+e^x\right)-\ln \left(1+e^0\right)\right]-e^{-2x}\left[e^x-\ln \left(1+e^x\right)-e^0+\ln \left(1+e^0\right)\right] \\ =e^{-x}\ln \left(1+e^x\right)-e^{-x}\ln 2-e^{-x}+e^{-2x}\ln \left(1+e^x\right)+e^{-2x}-e^{-2x}\ln 2.............\left(b\right) \end{gathered}$$

Then from (a) and (b), we can obtain the general solution of the given differential equation as

$$\begin{gathered} y=y_c+y_p \\ =k_1{e}^{-x}+k_2{e}^{-2x}+e^{-x}\ln \left(1+e^x\right)-e^{-x}\ln 2-e^{-x}+e^{-2x}\ln \left(1+e^x\right)+e^{-2x}-e^{-2x}\ln 2.......\left(c\right) \end{gathered}$$

Finally, we have to apply the initial condition of the given differential equation as the following:

We have to apply the point into $(x,y)=(0,0)$equation (c), then we have

$$\begin{gathered} 0=k_1{e}^0+k_2{e}^0+e^0\ln \left(1+e^0\right)-e^0\ln 2-e^0+e^0\ln \left(1+e^0\right)+e^0-e^0\ln 2 \\ 0=k_1+k_2+\ln 2-\ln 2-1+\ln 2+1-\ln 2 \\ {k}_1+k_2=0.............\left(d\right) \end{gathered}$$

After that, we have to take the first derivative for the general solution as

$$\begin{gathered} y^{\text{'}}=-k_1{e}^{-x}-2k_2{e}^{-2x}-e^{-x}\ln \left(1+e^x\right)+\frac{1}{1+e^x}+e^{-x}\ln 2+e^{-x} \\ -2e^{-2x}\ln \left(1+e^x\right)+\frac{e^{-x}}{1+e^x}-2e^{-2x}+2e^{-2x}\ln 2 \end{gathered}$$

We have to apply the point $(x,y)=(0,1)$in above equation, then we have

role="math" localid="1668060074076" $$\begin{gathered} 1=-k_1{e}^0-2k_2{e}^0-e^0\ln \left(1+e^0\right)+\frac{1}{1+e^0}+e^0\ln 2+e^0 \\ -2e^0\ln \left(1+e^0\right)+\frac{e^0}{1+e^0}-2e^0+2e^0\ln 2 \\ 1=-k_1-2k_2-\ln \left(2\right)+\frac{1}{2}+\ln 2+1-2\ln \left(2\right)+\frac{1}{2}-2+2\ln 2 \end{gathered}$$

$$\begin{gathered} 1=-k_1-2k_2 \\ {k}_1+2k_2=-1...............\left(e\right) \end{gathered}$$

Final proof

Then by solving these two equations (d) and (e), we find that

$k_1=1\text{and}{k}_2=-1$

Then, substitute with$k_1$ and$k_2$ into equation (c), then we obtain

$$\begin{gathered} y=e^{-x}-e^{-2x}+e^{-x}\ln \left(1+e^x\right)-e^{-x}\ln 2-e^{-x}+e^{-2x}\ln \left(1+e^x\right)+e^{-2x}-e^{-2x}\ln 2 \\ =\left[\ln \left(1+e^x\right)-\ln 2\right]e^{-x}+\left[\ln \left(1+e^x\right)-\ln 2\right]e^{-2x} \end{gathered}$$

is the solution of the given differential equation at the given conditions.