Question

Solve each differential equation by variation of parameters.

${\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{-}\mathit{y}\mathbf{=}\mathit{c}\mathit{o}\mathit{s}\mathit{h}\mathit{x}$

Short answer

$y=k_1{e}^x+k_2{e}^{-x}+\frac{1}{2}x\sinh x$

Step-by-step solution

Step 1: Determine the corresponding homogeneous equation

We have the second order differential equation

$y^{\text{'}\text{'}}-y=\cosh x,$

which is

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

as the form

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

and we have to get its general solution by the following technique :

First, we have to find the solution of the corresponding homogeneous differential equation as

$y^{\text{'}\text{'}}-y=0,$ ………. (1)

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

Differentiate the assumption with respect to x , then we have

$y^{\text{'}}=m{e}^{mx}$

and

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

After that, substitute with$y=e^{mx}$ and equation (2) into the differential equation (1), then we obtain

$\begin{array}{rcl}{m}^2{e}^{mx}-e^{mx}& =& 0\\ e^{mx}\left(m^2-1\right)& =& 0\end{array}$

Since$e^{mx}$ cannot be equal 0, the we have

$(m-1)(m+1)=0$

Then the roots are

$m_1=1\text{and}{m}_2=-1$

which are conjugate and complex.

Then we can obtain the solution of the homogeneous differential equation as

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

$y_h=c_1{e}^x+c_2{e}^x$ ……… (a)

Determine the particular solution for  differential equation

Second, now we have to find the particular solution of the given differential equation as the following technique:

Since we have $y_1=e^x$and $y_2=e^{-x}$, then we can obtain the wronskian as

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

After that, since we have the right side of the given differential equation $f\left(x\right)=\cosh x$, then we can obtain the wronskian $W_1$and $W_2$as

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

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

After we had obtained the wronskian $W_1$and $W_2$ , we have to obtain and as the following

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

and

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

After that, we have to do integration for both $u_1^{\text{'}}$and $u_2^{\text{'}}$neglecting integration constants to obtain$u_1$ and$u_2$ as the following

$\begin{array}{rcl}{u}_1& =& \int u_1^{\text{'}}dx\\ & =& \int \left(\frac{1+e^{2x}}{4}\right)dx\\ & =& \int \frac{1}{4}dx+\int \frac{1}{4}{e}^{2x}dx\\ & =& \frac{1}{4}x-\frac{1}{8}{e}^{-2x}\end{array}$

and

$\begin{array}{rcl}{u}_2& =& \int u_2^{\text{'}}dx\\ & =& \int -\left(\frac{e^{2x}+1}{4}\right)dx\\ & =& \int -\frac{1}{4}{e}^{2x}dx-\int \frac{1}{4}dx\\ & =& -\frac{1}{8}{e}^{2x}-\frac{1}{4}x\end{array}$

After that, using the values of role="math" localid="1664179698207" $u_1$ and $u_2$the functions role="math" localid="1664179759970" $y_1$and role="math" localid="1664179843371" $y_2$ , we can obtain the particular solution as

$\begin{array}{rcl}{y}_p& =& u_1{y}_1+u_2{y}_2\\ & =& \left(\frac{1}{4}x-\frac{1}{8}{e}^{-2x}\right)\times \left(e^x\right)+\left(-\frac{1}{8}{e}^{2x}-\frac{1}{4}x\right)\times \left(e^{-x}\right)\\ & =& \frac{1}{4}x{e}^x-\frac{1}{8}{e}^{-x}-\frac{1}{8}{e}^x-\frac{1}{4}x{e}^{-x}\\ & =& \frac{1}{2}x\left(\frac{1}{2}{e}^x-\frac{1}{2}{e}^{-x}\right)-\frac{1}{8}\left(e^x+e^{-x}\right)\\ y_p& =& \frac{1}{2}x\sinh x-\frac{1}{8}\left(e^x+e^{-x}\right)\end{array}$ …….. $y_p=\frac{1}{2}x\sinh x-\frac{1}{8}\left(e^x+e^{-x}\right)$(b)

Determine the general solution for given differential equation

Finally, from equations (a) and (b) we can obtain the general solution of the differential equation as

$\begin{array}{rcl}y& =& y_h+y_p\\ & =& c_1{e}^x+c_2{e}^{-x}+\frac{1}{2}x\sinh x-\frac{1}{8}\left(e^x+e^{-x}\right)\\ & =& \left(c_1-\frac{1}{8}\right)e^x+\left(c_2-\frac{1}{8}\right)e^{-x}+\frac{1}{2}x\sinh x\\ y& =& k_1{e}^x+k_2{e}^{-x}+\frac{1}{2}x\sinh x\end{array}$