Question

(d2y/dx2) - y = 4/(1+e^​x)

Answer 1

At first, we solve the equation: $y''-y=0$. To solve the equation, we substitute solution of the form:$g=e^{\lambda x}$ and receive the equation for $\lambda:$ $\lambda^2-1=0$. Solutions are: $\lambda=\pm1$. Thus, the solution of the homogenous equation $y''-y=0$ is: $g=c_1e^{x}+c_2e^{-x}$. The solution of the initial equation can be presented as: $y=g+f$, where $f$ is a solution of the initial equation. Substitute $f=ce^{x}$ with function $c(x)$ into the initial equation:

1). $c''e^{x}+2c'e^x+ce^x-ce^x=\frac{4}{1+e^x}$. Do the change of variables: $f_1=c'$ in the equation: $f_1'e^{x}+2f_1e^x=\frac{4}{1+e^x}$ .Solve another homogeneous equation: $f_1'=-2f_1$. The solution is: $f_1=b_1e^{-2x}$. Suppose that $b_1$ is a function. Substitute the latter expression into equation $f_1'e^{x}+2f_1e^x=\frac{4}{1+e^x}$ and receive: $b_1'=\frac{4e^{x}}{1+e^x}$ Compute the integral to find $f_1$: $b_1=4\int\frac{d(e^x)}{1+e^x}=4(ln(1+e^x)+C_1)$. $f_1=4e^{-2x}ln(1+e^x)+e^{-2x}C_1$. $c=\int fdx=\int(4e^{-2x}ln(1+e^x)+4e^{-2x}C_1)dx=I_1+I_2$, where $I_1=4\int e^{-2x}ln(1+e^x)dx$, $I_2=4\int e^{-2x}C_1dx$. Compute the integral $I_1$ :

$I_1=4\int e^{-2x}ln(1+e^x)dx$. Make the change of variables: $z=e^{x}$. The integral becomes: $I_1=4\int z^{-3}ln(1+z)dz=-2{z^{-2}}ln(1+z)+2\int \frac{z^{-2}}{1+z}dz=-2{e^{-2x}}ln(1+e^x)+2(-e^{-x}+ln(1+e^x)-x)+a_1$

$I_2=-2C_1e^{-2x}+a_2$

Thus, $c=-2{e^{-2x}}ln(1+e^x)+2(-e^{-x}+ln(1+e^x)-x)+a_1-2C_1e^{-2x}+a_2$ and, respectively:

$f=-2+2(e^{x}-e^{-x})ln(1+e^{x})-2xe^x-2C_1e^{-x}+Ke^x$. Last two terms can be absorbed by $g$ and $y=c_1e^{x}+c_2e^{-x}-2+2(e^{x}-e^{-x})ln(1+e^{x})-2xe^x$ This is the solution of the problem.