Question

In Problems 19–22 solve each differential equation by variation of parameters, subject to the initial conditions

$\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathit{y}\mathbf{\text{'}}\mathbf{-}\mathbf{8}\mathit{y}\mathbf{=}\mathbf{2}{\mathit{e}}^{\mathbf{-}\mathbf{2}\mathbf{x}}\mathbf{-}{\mathit{e}}^{\mathbf{-}\mathbf{x}}$

Short answer

$y=\frac{25}{36}{e}^{2x}+\frac{4}{9}{e}^{-4x}-\frac{1}{4}{e}^{-2x}+\frac{1}{9}{e}^{-x}$

Step-by-step solution

Apply the differential equation with initial conditions

The given Equation is a second order differential equation

$y\text{'}\text{'}+2y\text{'}-8y=2e^{-2x}-e^{-x}$

In the Form as

$y\text{'}\text{'}+p\left(x\right)y\text{'}+q\left(x\right)y=f\left(x\right)$,

With initial conditions $y\left(0\right)=1,y\text{'}\left(0\right)=0$.

To Obtain General Solution using the following technique,

Finding the solution of the corresponding homogeneous differential equation as

$y\text{'}\text{'}+2y\text{'}-8y=0······\left(1\right)$

Assuming that $y=e^{mx}$is a solution for the homogeneous differential equation, then follow the following technique:

Differentiate with respect to x, then we have

$$\begin{gathered} y\text{'}=m{e}^{mx}······\left(2\right) \\ y\text{'}\text{'}=m^2{e}^{mx}······\left(3\right) \end{gathered}$$

After that, Substitute with y = e^mxand equation (2) and (3) into the differential equation (1), then we obtain

$$\begin{gathered} m^2{e}^{mx}+2m{e}^{mx}-8e^{mx}=0 \\ {e}^{mx}\left(m^2+2m-8\right)=0 \end{gathered}$$

cannot be 0 so,

$$\begin{gathered} m^2+2m-8=0 \\ (m-2)(m+4)=0 \end{gathered}$$

Use wronskian with roots

The Roots which are real and distinct are,

$m_1=2and{m}_2=-4$

Obtaining the solution of the homogeneous differential Equation as

$y_h=c_1{e}^{2x}+c_2{e}^{-4x}······\left(\text{a}\right)$

To find the particular solution y_pof the given differential equation as the following technique:

Since we have $y_1=e^{2x}\hspace{0.17em}and{y}_2=e^{-4x}$, Hence we can obtain the Wronskian as

$$\begin{gathered} W\left(y_1,y_2\right)=\left|Y_1{Y}_2 \\ Y{\text{'}}_{1}Y{\text{'}}_2\right| \\ =\left|e^{2x}{e}^{-4x} \\ 2e^{2x}-4e^{-4x}\right| \\ =\left(e^{2x}\right)\times \left(-4e^{-4x}\right)-\left(e^{-4x}\right)\times \left(2e^{2x}\right) \\ =-4e^{-2x}-2e^{-2x} \\ =-6e^{-2x} \end{gathered}$$

Apply wronskian to solve the differential equation

Take the right side of the differential equation, $f\left(x\right)=2e^{-2x}-e^{-x}$and to find the Wronskian W₁ and W₂ as,

role="math" localid="1668415967169" $$\begin{gathered} W_1=\left|0y_2 \\ f\left(x\right)y_2^{\text{'}}\right| \\ =\left|0e^{-4x} \\ 2e^{-2x}-e^{-x}-4e^{-4x}\right| \\ =0-\left(e^{-4x}\right)\times \left(2e^{-2x}-e^{-x}\right) \\ =-2e^{-6x}+e^{-5x} \\ {W}_2=\left|y_1{y}_2 \\ {y}_1^{\text{'}}f\left(x\right)\right| \\ =\left|e^{2x}0 \\ 2e^{2x}2e^{-2x}-e^{-x}\right| \\ =\left(e^{2x}\right)\times \left(2e^{-2x}-e^{-x}\right)-0 \\ =2-e^x \\ {u}_1\text{'}=\frac{W_1}{W\left(y_1,y_2\right)} \\ =\frac{-2e^{-6x}+e^{-5x}}{-6e^{-2x}} \\ =\frac{1}{3}{e}^{-4x}-\frac{1}{6}{e}^{-3x} \\ {u}_2\text{'}=\frac{W_1}{W\left(y_1,y_2\right)} \\ =\frac{2-e^x}{-6e^{-2x}} \\ =-\frac{1}{3}{e}^{2x}+\frac{1}{6}{e}^{3x} \end{gathered}$$

We have to do Integration method for both u₁' and u'₂neglecting integration constants u₁ and u₂ as the following,

$$\begin{gathered} u_1=\int u_1\text{'}dx \\ =\int \left(\frac{1}{3}{e}^{-4x}-\frac{1}{6}{e}^{-3x}\right)dx \\ =\frac{1}{3}{e}^{-4x}\times -\frac{1}{4}-\frac{1}{6}{e}^{-3x}\times -\frac{1}{3} \\ =-\frac{1}{12}{e}^{-4x}+\frac{1}{18}{e}^{-3x} \\ {u}_2=\int u_2\text{'}dx \\ =\int \left(-\frac{1}{3}{e}^{2x}+\frac{1}{6}{e}^{3x}\right)dx \\ =-\frac{1}{3}{e}^{2x}\times \frac{1}{2}+\frac{1}{6}{e}^{3x}\times \frac{1}{3} \\ =-\frac{1}{6}{e}^{2x}+\frac{1}{18}{e}^{3x} \end{gathered}$$

Apply the initial conditions

Using u₁ and u₂the functions we can obtain the particular solution as,

$$\begin{gathered} y_p=u_1{y}_1+u_2{y}_2 \\ =\left(-\frac{1}{12}{e}^{-4x}+\frac{1}{18}{e}^{-3x}\right)\times \left(e^{2x}\right)+\left(-\frac{1}{6}{e}^{2x}+\frac{1}{18}{e}^{3x}\right)\times \left(e^{-4x}\right) \\ =-\frac{1}{12}{e}^{-2x}+\frac{1}{18}{e}^{-x}-\frac{1}{6}{e}^{-2x}+\frac{1}{18}{e}^{-x} \\ =-\frac{1}{4}{e}^{-2x}+\frac{1}{9}{e}^{-x}······\left(\text{b}\right) \end{gathered}$$

From equation (a) and (b) we can obtain the general solution

$$\begin{gathered} y\text{'}\text{'}+2y\text{'}-8y=2e^{-2x}-e^{-x} \\ y=y_h+y_p \\ =c_1{e}^{2x}+c_2{e}^{-4x}-\frac{1}{4}{e}^{-2x}+\frac{1}{9}{e}^{-x}······\left(\text{c}\right) \end{gathered}$$

Applying Initial conditions for the given differential equation, we have to apply the point into $(x,y)=(0,1)$(c), then we have

$$\begin{gathered} 1=c_1{e}^0+c_2{e}^0-\frac{1}{4}{e}^0+\frac{1}{9}{e}^0 \\ {c}_1+c_2=\frac{41}{36}······\left(d\right) \end{gathered}$$

We have to take the first derivative for the general solution shown in (c) as

$y\text{'}=2c_1{e}^{2x}-4c_2{e}^{-4x}+\frac{1}{2}{e}^{-2x}-\frac{1}{9}{e}^{-x}$

Then apply the point $\left(x,y\text{'}\right)=(0,0)$we have,

$$\begin{gathered} 0=2c_1{e}^0-4c_2{e}^0+\frac{1}{2}{e}^0-\frac{1}{9}{e}^0 \\ 2c_1-4c_2=-\frac{7}{18}······\left(e\right) \end{gathered}$$

By solving the equation (d) and (e), we find that

$c_1=\frac{25}{36}and{c}_2=\frac{4}{9}$

Substitute the value of into (c) then we obtain,

$y=\frac{25}{36}{e}^{2x}+\frac{4}{9}{e}^{-4x}-\frac{1}{4}{e}^{-2x}+\frac{1}{9}{e}^{-x}$.

Hence the solution is role="math" localid="1668416978353" $\mathit{y}\mathbf{=}\frac{\mathbf{25}}{\mathbf{36}}{\mathit{e}}^{\mathbf{2}\mathbf{x}}\mathbf{+}\frac{\mathbf{4}}{\mathbf{9}}{\mathit{e}}^{\mathbf{-}\mathbf{4}\mathbf{x}}\mathbf{-}\frac{\mathbf{1}}{\mathbf{4}}{\mathit{e}}^{\mathbf{-}\mathbf{2}\mathbf{x}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{9}}{\mathit{e}}^{\mathbf{-}\mathbf{x}}$