Question

Solve each differential equation by variation of parameters

$\mathbf{4}{\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{-}\mathit{y}\mathbf{=}{\mathit{e}}^{\frac{\mathbf{1}}{\mathbf{2}}\mathbf{t}}\mathbf{+}\mathbf{3}$

Short answer

$y=c_1{e}^{\frac{1}{2}t}+c_2{e}^{\frac{1}{2}t}+\frac{1}{4}t{e}^{\frac{1}{2}t}-\frac{1}{8}{e}^{\frac{3}{2}t}-3$

Step-by-step solution

Step 1: Determine the corresponding homogeneous equation

We have the second order differential equation

$4y^{\text{'}\text{'}}-y=e^{\frac{1}{2}t}+3$

which is

$y^{\text{'}\text{'}}-\frac{1}{4}y=\frac{e^{\frac{1}{2}t}+3}{4}$

as the form

$y^{\text{'}\text{'}}+p\left(t\right)y^{\text{'}}+q\left(t\right)y=f\left(t\right)$

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

$y^{\text{'}\text{'}}-\frac{1}{4}y=0$ …….. (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

$\begin{array}{rcl}{m}^2{e}^{mt}-e^{mt}& =& 0\\ e^{mt}\left(m^2-\frac{1}{4}\right)& =& 0\\ \left(m-\frac{1}{2}\right)\left(m+\frac{1}{2}\right)& =& 0\end{array}$

Then the roots are

$m_1=\frac{1}{2}\text{and}{m}_2=-\frac{1}{2}$

which are real and distinct

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

$\begin{array}{rcl}{y}_h& =& c_1{y}_1+c_2{y}_2\\ & =& c_1{e}^{\frac{1}{2}t}+c_2{e}^{-\frac{1}{2}t}\end{array}$

$y_h=c_1{e}^{\frac{1}{2}t}+c_2{e}^{-\frac{1}{2}t}$ ……. (b)

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$y_1=e^{\frac{1}{2}t}$and$y_2=e^{-\frac{1}{2}t}$, then we can obtain the wronskian as

$\begin{array}{rcl}W\left(y_1,y_2\right)& =& \left|\begin{array}{cc}{y}_1& y_2\\ y{\text{'}}_1& y{\text{'}}_2\end{array}\right|\\ & =& \left|\begin{array}{cc}{e}^{\frac{1}{2}t}& e^{\frac{1}{2}t}\\ \frac{e^{\frac{1}{2}t}}{2}& -\frac{e^{\frac{1}{2}t}}{2}\end{array}\right|\\ & =& \left(e^{\frac{1}{2}t}\right)\times \left(-\frac{e^{-\frac{1}{2}t}}{2}\right)-\left(e^{\frac{1}{2}t}\right)\times \left(-\frac{e^{\frac{1}{2}t}}{2}\right)\\ & =& -\frac{1}{2}{e}^0-\frac{1}{2}{e}^0\\ & =& -1\end{array}$

After that, since we have the right side of the given differential equation $f\left(t\right)=\frac{e^{\frac{1}{2}t}+3}{4}$, then we can obtain the wronskian $W_{1}and{W}_2$as

$\begin{array}{rcl}{W}_1& =& \left|\begin{array}{cc}0& y_2\\ f\left(t\right)& y{\text{'}}_2\end{array}\right|\\ & =& \left|\begin{array}{cc}0& e^{\frac{1}{2}t}\\ \frac{e^{\frac{1}{2}t}+3}{4}& -\frac{e^{-\frac{1}{2}t}}{2}\end{array}\right|\\ & =& 0-e^{\frac{1}{2}t}\times \left(\frac{e^{\frac{1}{2}t}+3}{4}\right)\in \\ & =& -\frac{1}{4}{e}^{\frac{1}{2}t}\times e^{\frac{1}{2}t}-\frac{3}{4}\times e^{\frac{1}{2}t}\\ & =& -\frac{1}{4}{e}^0-\frac{3}{4}{e}^{\frac{1}{2}t}\\ & =& -\frac{1}{4}\left(3e^{\frac{1}{2}t}+1\right)\\ & & \end{array}$

$\begin{array}{rcl}{W}_2& =& \left|\begin{array}{cc}{y}_1& 0\\ y{\text{'}}_1& f\left(t\right)\end{array}\right|\\ & =& \left|\begin{array}{cc}{e}^{\frac{1}{2}t}& 0\\ \frac{e^{\frac{1}{2}t}}{2}& \frac{e^{\frac{1}{2}t}+3}{4}\end{array}\right|\\ & =& e^{\frac{1}{2}t}\times \left(\frac{e^{\frac{1}{2}t}+3}{4}\right)-0\\ & =& \frac{1}{4}{e}^{\frac{1}{2}t}\times e^{\frac{1}{2}t}+\frac{3}{4}\times e^{\frac{1}{2}t}\\ & =& \frac{1}{4}{e}^{2t}+\frac{3}{4}{e}^{\frac{1}{2}t}\end{array}$

After we had obtained the wronskian $W_{1}and{W}_2$, we have to obtain $u{\text{'}}_{1}andu{\text{'}}_2$as the following

$\begin{array}{rcl}{u}_1^{\text{'}}& =& \frac{W_1}{W\left(y_1,y_2\right)}\\ & =& \frac{-\frac{1}{4}\left(3e^{\frac{1}{2}t}+1\right)}{-1}\\ & =& \frac{1}{4}\left(3e^{\frac{1}{2}t}+1\right)\end{array}$

and

$\begin{array}{rcl}{u}_2^{\text{'}}& =& \frac{W_2}{W\left(y_1,y_2\right)}\\ & =& \frac{\frac{1}{4}\left(e^{2t}+3e^{\frac{1}{2}t}\right)}{-1}\\ & =& -\frac{1}{4}\left(e^{2t}-3e^{\frac{1}{2}t}\right)\\ & =& -\frac{1}{4}{e}^{2t}\times \frac{1}{2}-\frac{3}{4}{e}^{\frac{1}{2}t}\times \left(2\right)\\ & =& -\frac{1}{8}{e}^{2t}-\frac{3}{2}{e}^{\frac{1}{2}t}\end{array}$

Determine the general solution for given differential equation

After that, using the values of$u_{1}and{u}_{2}thefunctions{y}_{1}and{y}_2,$ , we can obtain the particular solution as

$\begin{array}{rcl}{y}_p& =& u_1{y}_1+u_2{y}_2\\ & =& \left(\frac{1}{4}t-\frac{3}{2}{e}^{\frac{1}{2}t}\right)\times \left(e^{\frac{1}{2}t}\right)+\left(-\frac{1}{8}{e}^{2t}-\frac{3}{2}{e}^{\frac{1}{2}t}\right)\times \left(e^{-\frac{1}{2}t}\right)\\ & =& \frac{1}{4}t{e}^{\frac{1}{2}t}-\frac{3}{2}-\frac{1}{8}{e}^{\frac{3}{2}t}-\frac{3}{2}\\ y_p& =& \frac{1}{4}t{e}^{\frac{1}{2}t}-\frac{1}{8}{e}^{\frac{3}{2}t}-3\end{array}$

$y_p=\frac{1}{4}t{e}^{\frac{1}{2}t}-\frac{1}{8}{e}^{\frac{3}{2}t}-3$ ……… (b)

Finally, from equations (a) and (b) we can obtain the general solution of the differential$y^{\text{'}\text{'}}-y=\cosh x$as

$$\begin{gathered} y=y_h+y_p \\ y=c_1{e}^{\frac{1}{2}t}+c_2{e}^{\frac{1}{2}t}+\frac{1}{4}t{e}^{\frac{1}{2}t}-\frac{1}{8}{e}^{\frac{3}{2}t}-3 \end{gathered}$$