Question

Solve each differential equation by variation of parameters.

${\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{+}\mathbf{3}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{+}\mathbf{2}\mathit{y}\mathbf{=}\mathit{s}\mathit{i}\mathit{n}{\mathit{e}}^{\mathbf{x}}$

Short answer

$c_1{e}^{-x}+c_2{e}^{-2x}-e^{-2x}\sin e^x$

Step-by-step solution

Note the given data

Given the differential equation$y^{\text{'}\text{'}}+3y^{\text{'}}+2y=\sin e^x,$ .

We need to find the general solution of the given differential equation

Finding the solution of the homogeneous solution

Finding the solution of the homogeneous equation

$y^{\text{'}\text{'}}+3y^{\text{'}}+2y=0$…………………………………………………………..(1)

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

Differentiate the above equation with respect to x , we get

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

Again, differentiate the above equation with respect to x, we get

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

Substitute the value$y=e^{mx}$ and equation (3) in differential equation (1), we get

$\begin{array}{rcl}{m}^2{e}^{mx}+3m{e}^{mx}+2e^{mx}& =& 0\\ e^{mt}\left(m^2+3m+2\right)& =& 0\end{array}$

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

$m=\pm i$

Therefore, the roots are complex and conjugate.

Then the solution of a homogeneous solution is

$y_h=c_1{y}_1+c_2{y}_2$

$=c_1{e}^{-x}+c_2{e}^{-2x}$………………………………………………………………..(4)

Finding the Wronskian from left side

We need to find the particular solution

Since we have$y_1=e^{-z}$and$y_2=e^{-2x}$, then we can obtain the wronskian as

$\begin{array}{rcl}W\left(y_1,y_2\right)& =& \left|\begin{array}{cc}{y}_1& y_2\\ y{\text{'}}_1& y{\text{'}}_2\end{array}\right|\\ & =& \left|\begin{array}{cc}{e}^{-x}& e^{2-x}\\ 1e^{-x}& -e^{-2x}\end{array}\right|\\ & =& \left(e^{-x}\right)\times \left(-2e^{-2x}\right)-\left(e^{-2x}\right)\times \left(-e^{-x}\right)\\ & =& \left(-2e^{-x-2x}\right)-\left(-e^{-2x-x}\right)\\ & =& -2e^{-3x}+e^{-3x}\\ & =& -e^{-3x}\\ & & \end{array}$

Finding the Wronskian from right side of given differential question

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

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

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

And

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

Finding the value of u1 and u2

Now, we find the value of

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

And

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

By integration, we get

$\begin{array}{rcl}{u}_1& =& \int u_1^{\text{'}}dx\\ & =& \int e^x\sin e^{x}dx\end{array}$

And

$\begin{array}{rcl}{u}_1& =& \int u_1^{\text{'}}dx\\ & =& \int \sin vdv\\ & =& -\cos v\\ & =& -\cos e^z\end{array}$

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

$\begin{array}{rcl}{y}_p& =& u_1{y}_1+u_2{y}_2\\ & =& \left(-\cos e^x\right)\times \left(e^{-x}\right)+\left(e^x\cos e^x-\sin e^x\right)\times \left(e^{-2x}\right)\\ & =& -e^{-x}\cos e^x+e^{-x}\cos e^x-e^{-2x}\sin e^x\\ & =& -e^{-2x}\sin e^x\end{array}$

Now, we find the general solution

$\begin{array}{rcl}y& =& y_h+y_p\\ & =& c_1{e}^{-x}+c_2{e}^{-2x}-e^{-2x}\sin e^x\end{array}$