Question

Solve each differential equation by variation of parameters.

${\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{-}\mathit{y}\mathbf{=}\mathit{s}\mathit{i}\mathit{n}\mathit{h}\mathbf{2}\mathit{x}$

Short answer

$y=c_1{e}^x+c_2{e}^{-x}+\frac{1}{3}\sinh 2x$

Step-by-step solution

Given data

Given the differential equation $y^{\text{'}\text{'}}-y=\sinh 2x$.

The objective is to find the general solution of the given differential equation

Finding the solution of the homogeneous solution

Finding the solution of the homogeneous equation

$y^{\text{'}\text{'}}-y=0$ ...(1)

Let $y=e^{mx}$ be a solution of the given homogeneous solution

Differentiate the above equation with respect to x gives:

$y^{\text{'}}=m{e}^{mx}$ …(2)

Again differentiate the above equation with respect to x gives:

$y^{\text{'}\text{'}}=m^2{e}^{mx}$ ...(3)

Substitute the value and equation (3) in differential equation (1), we get

$$\begin{gathered} m^2{e}^{mx}-e^{mx}=0 \\ \Rightarrow e^{mx}\left(m^2-1\right)=0 \end{gathered}$$

Since $e^{mx}\ne 0$, then the roots are

$$\begin{gathered} \left(m-1\right)\left(m+1\right)=0 \\ {m}_1=1,m_2=-1 \end{gathered}$$

Therefore, the roots are real and distinct.

Then the solution of a homogeneous solution is

$\begin{array}{rcl}{y}_h& =& c_1{y}_1+c_2{y}_2\\ & =& c_1{e}^x+c_2{e}^{-x}\end{array}$ ...(4)

Finding the Wronskian from left side

Find the particular solution .

We have $y_1=e^x$ and $y_2=e^{-x}$. So,

$\begin{array}{rcl}W\left(y_1,y_2\right)& =& \left|\begin{array}{cc}{y}_1& y_2\\ y{\text{'}}_1& y{\text{'}}_2\end{array}\right|\\ & =& \left|\begin{array}{cc}{e}^x& e^{-x}\\ e^x& -e^{-x}\end{array}\right|\\ & =& \left(e^x\right)\times \left(-e^{-x}\right)-\left(e^{-x}\right)\times \left(e^x\right)\\ & =& -e^{x-x}-\left(e^{x-x}\right)\\ & =& -1-1\\ & =& -2\end{array}$

Finding the Wronskian from right side of given differential question

Now, find the Wronskian $W_1\text{and}{W}_2$ of right side of the differential equation

$f\left(x\right)=\cosh x$

$\begin{array}{rcl}{W}_1& =& \left|\begin{array}{cc}0& y_2\\ f\left(x\right)& y{\text{'}}_2\end{array}\right|\\ & =& \left|\begin{array}{cc}0& e^{-x}\\ \frac{1}{2}{e}^{2x}-\frac{1}{2}{e}^{-2x}& -e^{-x}\end{array}\right|\\ & =& 0-e^{-x}\left(\frac{1}{2}{e}^{2x}-\frac{1}{2}{e}^{-2x}\right)\\ & =& -\frac{1}{2}{e}^{2x}\times e^{-x}-\frac{1}{2}{e}^{-2x}\times e^{-x}\\ & =& -\frac{1}{2}-\frac{1}{2}{e}^{-3x}\\ & =& \frac{1}{2}\left(e^{-3x}-e^x\right)\end{array}$

And

$\begin{array}{rcl}{W}_2& =& \left|\begin{array}{cc}0& y_2\\ f\left(x\right)& y_2\text{'}\end{array}\right|\\ & =& \left|\begin{array}{cc}{e}^x& 0\\ e^x& \frac{1}{2}{e}^{2x}-\frac{1}{2}{e}^{-2x}\end{array}\right|\\ & =& e^x\times \left(\frac{1}{2}{e}^{2x}-\frac{1}{2}{e}^{-2x}\right)-0\\ & =& \frac{1}{2}{e}^{2x}\times e^x-\frac{1}{2}{e}^{-2x}\times e^x\\ & =& \frac{1}{2}{e}^{3x}-\frac{1}{2}{e}^{-x}\\ & =& \frac{1}{2}\left(e^{3x}-e^{-x}\right)\end{array}$

Finding the value of u1 and   u2

Now, find the value of

$\begin{array}{rcl}{u}_1^{\text{'}}& =& \frac{W_1}{W\left(y_1,y_2\right)}\\ & =& \frac{\frac{1}{2}\left(e^{-3x}-e^x\right)}{-2}\\ & =& \frac{\left(e^x-e^{-3x}\right)}{4}\end{array}$

And

$\begin{array}{rcl}{u}_2^{\text{'}}& =& \frac{W_2}{W\left(y_1,y_2\right)}\\ & =& \frac{\frac{1}{2}\left(e^{3x}-e^{-x}\right)}{-2}\\ & =& \frac{e^{-x}-e^{-3x}}{4}\end{array}$

By integration, we get

$\begin{array}{rcl}{u}_1& =& \int u_1^{\text{'}}dx\\ & =& \int \left(\frac{e^x-e^{-3x}}{4}\right)dx\\ & =& \int \frac{1}{4}{e}^{x}dx-\int \left(\frac{e^{-3x}}{4}\right)dx\\ & =& \frac{1}{4}{e}^x+\frac{e^{-3x}}{12}\end{array}$

And

$\begin{array}{rcl}{u}_2& =& \int u_2^{\text{'}}dx\\ & =& \int -\left(\frac{e^{-x}-e^{-3x}}{4}\right)dx\\ & =& \int \left(\frac{e^{-x}}{4}\right)dx-\int \left(\frac{1}{4}\right)e^{-3x}dx\\ & =& -\frac{e^{-x}}{4}-\frac{1}{12}{e}^{3x}\end{array}$

Finding the required solution and we get

Substitute all the values of$u_1\text{and}{u}_2\text{and the functions}{y}_1\text{and}{y}_2$ , we get the particular solution

$\begin{array}{rcl}{\text{y}}_p& =& u_1{y}_1+u_2{y}_2\\ & =& \left(\frac{{\displaystyle 1}}{{\displaystyle 4}}{e}^x+\frac{{\displaystyle e^{-3x}}}{{\displaystyle 12}}\right)\times e^x+\left(-\frac{{\displaystyle 1}}{{\displaystyle 4}}{e}^{-x}-\frac{{\displaystyle e^{3x}}}{{\displaystyle 12}}\right)\times e^{-x}\\ & =& \left(\frac{{\displaystyle 1}}{{\displaystyle 4}}{e}^{2x}+\frac{{\displaystyle e^{-2x}}}{{\displaystyle 12}}-\frac{{\displaystyle 1}}{{\displaystyle 4}}{e}^{-2x}-\frac{{\displaystyle 1}}{{\displaystyle 12}}{e}^{2x}\right)\\ & =& \left(\frac{{\displaystyle 3-1}}{{\displaystyle 12}}\right)e^{2x}+\left(\frac{{\displaystyle 1-3}}{{\displaystyle 12}}\right)\left(e^{-2x}\right)\\ & =& \frac{{\displaystyle 1}}{{\displaystyle 6}}\left(e^{2x}-e^{-2x}\right)\\ & =& \frac{{\displaystyle 1}}{{\displaystyle 3}}\left(\frac{{\displaystyle e^{2x}-e^{-2x}}}{{\displaystyle 2}}\right)\\ & =& \frac{{\displaystyle 1}}{{\displaystyle 3}}\sinh 2x\end{array}$

Now, the general solution is

$$\begin{gathered} y=y_c+y_p \\ y=c_1{e}^x+c_2{e}^{-x}+\frac{1}{3}\sinh 2x \end{gathered}$$