Question

Solve, using the method of variation of parameters
(d2y/dx2) - y = 2/(1+e^​x)

Answer 1

Answer on Question 68963 - Math - Differential Equations

Solve, using the method of variation of parameters $\frac{d^2y}{dx^2} - y = \frac{2}{1 + e^x}$.

Solution:

Let us first solve the corresponding homogenous linear differential equation

$$\frac {d ^ {2} y}{d x ^ {2}} - y = 0.$$

The characteristic equation $\lambda^2 - 1 = 0$ has two roots $\lambda_1 = -1$ and $\lambda_2 = 1$. Consequently, the pair of functions $e^{-x}$ and $e^x$ is a fundamental system of solutions and therefore the general solution has the form

$$y = C _ {1} e ^ {- x} + C _ {2} e ^ {x},$$

where $C_1$ and $C_2$ are arbitrary real constants.

By the method of variation of parameters, we look for a partial solution of the non-homogenous equation in the form

$$y _ {*} = \alpha_ {1} (x) e ^ {- x} + \alpha_ {2} (x) e ^ {x}$$

with unknown functions $\alpha_{1}$ and $\alpha_{2}$. The derivatives of $\alpha_{1}$, $\alpha_{2}$ can be found as a solution of the system

$$\left( \begin{array}{cc} e ^ {- x} & e ^ {x} \\ - e ^ {- x} & e ^ {x} \end{array} \right) \left( \begin{array}{c} \alpha_ {1} ^ {\prime} \\ \alpha_ {2} ^ {\prime} \end{array} \right) = \left( \begin{array}{c} 0 \\ \frac {2}{1 + e ^ {x}} \end{array} \right).$$

Using Cramer's rule we solve the system

$$\Delta (x) = \left| \begin{array}{cc} e ^ {- x} & e ^ {x} \\ - e ^ {- x} & e ^ {x} \end{array} \right| = e ^ {- x} e ^ {x} + e ^ {- x} e ^ {x} = 2;$$ $$\Delta_ {1} (x) = \left| \begin{array}{cc} 0 & e ^ {x} \\ \frac {2}{1 + e ^ {x}} & e ^ {x} \end{array} \right| = - \frac {2 e ^ {x}}{1 + e ^ {x}},$$ $$\Delta_ {2} (x) = \left| \begin{array}{cc} e ^ {- x} & 0 \\ - e ^ {- x} & \frac {2}{1 + e ^ {x}} \end{array} \right| = \frac {2 e ^ {- x}}{1 + e ^ {x}};$$ $$\alpha_ {1} ^ {\prime} (x) = \frac {\Delta_ {1} (x)}{\Delta (x)} = - \frac {e ^ {x}}{1 + e ^ {x}}, \quad \alpha_ {2} ^ {\prime} (x) = \frac {\Delta_ {2} (x)}{\Delta (x)} = \frac {e ^ {- x}}{1 + e ^ {x}}.$$

Then we have

$$\begin{array}{l} \alpha_ {1} (x) = - \int \frac {e ^ {x}}{1 + e ^ {x}} d x = \left[ \begin{array}{c} e ^ {x} = t \\ x = \ln t \\ d x = \frac {1}{t} d t \end{array} \right] = - \int \frac {d t}{1 + t} = - \ln | t + 1 | \\ = - \ln (1 + e ^ {x}). \end{array}$$

Next,

$$\alpha_2(x) = \int \frac{e^{-x}}{1 + e^x} dx = - \int \frac{de^{-x}}{1 + e^x} = \left[ e^{-x} = t \right] = - \int \frac{dt}{1 + \frac{1}{t}} = - \int \frac{tdt}{t + 1} = - \int \left(1 - \frac{1}{t + 1}\right) dt = - t + \ln |t + 1| = - e^{-x} + \ln |e^{-x} + 1| = - e^{-x} + \ln (1 + e^x) - x.$$

Substituting $\alpha_1$, $\alpha_2$ into (1) gives the partial solution of the non-homogenous equation

$$y_* = (e^x - e^{-x}) \ln(1 + e^x) - x e^x - 1.$$

Finally we have the general solution of the non-homogenous equation

$$y = C_1 e^{-x} + C_2 e^x + (e^x - e^{-x}) \ln(1 + e^x) - x e^x - 1.$$

Answer: $y = C_1 e^{-x} + C_2 e^x + (e^x - e^{-x}) \ln(1 + e^x) - x e^x - 1$.

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