Question

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Answer 1

Answer on Question #60152 – Math – Differential Equations

Question

Use the method of variation of parameters to find a particular solution of the following differential equation

$$y'' - y = \frac{e^x}{e^x + 1}$$

Solve the given differential equation

Solution

First we need to find the general solution to the homogeneous equation

$$y'' - y = 0.$$

The characteristic equation for this ODE is

$$\lambda^2 - 1 = 0 \Rightarrow \lambda = \pm 1.$$

Hence the general solution is

$$y_g(x) = c_1 e^x + c_2 e^{-x}.$$

We shall use the method of variation of parameters to find a solution of the form

$$y_g(x) = y_1(x) e^x + y_2(x) e^{-x}.$$

The variation of parameters formula gives the system

$$\left\{ \begin{array}{c} y_1'(x) e^x + y_2'(x) e^{-x} = 0 \\ y_1'(x) e^x - y_2'(x) e^{-x} = \frac{e^x}{e^x + 1} \end{array} \right.$$

Adding the equations obtain

$$2 y_1'(x) e^x = \frac{e^x}{e^x + 1}$$ $$y_1'(x) = \frac{1}{2(e^x + 1)}$$

Integrating both sides with respect to $x$

$$y_1(x) = \int \frac{dx}{2(e^x + 1)} = \frac{1}{2}(x - \ln(e^x + 1)) + C_3.$$

Using the first equation $y_1'(x) e^x + y_2'(x) e^{-x} = 0$ of the system and plugging $y_1'(x) = \frac{1}{2(e^x + 1)}$, gives

$$y_2'(x) = - y_1'(x) e^{2x} = -\frac{e^{2x}}{2(e^x + 1)}$$

Integrating both sides with respect to $x$

$$y _ {2} (x) = \int \left(- \frac {e ^ {2 x}}{2 (e ^ {x} + 1)}\right) d x = - \frac {1}{2} (e ^ {x} - \ln (e ^ {x} + 1)) + C _ {4}.$$

The general solution is

$$\begin{array}{l} y _ {g} (x) = \frac {e ^ {x}}{2} (x - \ln (e ^ {x} + 1) + C _ {3}) - \frac {e ^ {- x}}{2} (e ^ {x} - \ln (e ^ {x} + 1) + C _ {4}) = \\ = C e ^ {x} + D e ^ {- x} + \frac {e ^ {x}}{2} (x - \ln (e ^ {x} + 1)) - \frac {e ^ {- x}}{2} (e ^ {x} - \ln (e ^ {x} + 1)). \\ \end{array}$$

Set $C = 0, D = 0$ .

Hence a particular solution is given by

$$y _ {p} (x) = \frac {e ^ {x}}{2} (x - \ln (e ^ {x} + 1)) - \frac {e ^ {- x}}{2} (e ^ {x} - \ln (e ^ {x} + 1)).$$

Answer:

$$y _ {p} (x) = \frac {e ^ {x}}{2} (x - \ln (e ^ {x} + 1)) - \frac {e ^ {- x}}{2} (e ^ {x} - \ln (e ^ {x} + 1));$$ $$y _ {g} (x) = C e ^ {x} + D e ^ {- x} + \frac {e ^ {x}}{2} (x - \ln (e ^ {x} + 1)) - \frac {e ^ {- x}}{2} (e ^ {x} - \ln (e ^ {x} + 1)).$$

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