Question

In problems 1-18 solve each differential equation by variation of parameters.

${\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{+}\mathit{y}\mathbf{=}\mathit{s}\mathit{e}\mathit{c}\mathit{x}$

Short answer

$y=c_1\cos x+c_2\sin x+\cos x\ln \left(\cos x\right)+x\sin x$

Step-by-step solution

Note the given data

Given the differential equation$y^{\text{'}\text{'}}+y=\sec x$ .

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.

$y^{\text{'}\text{'}}+y=0$ ...(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

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

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

$m=\pm i$

Therefore, the roots are complex and conjugate.

Then the solution of a homogeneous solution is

$\begin{array}{rcl}{y}_c& =& c_1{y}_1+c_2{y}_2\\ & =& c_1\cos x+c_2\sin x\end{array}$ ...(4)

Finding the Wronskian from left side

We need to find the particular solution .

We have $y_1=\cos x$and$y_2=\sin x$.

$\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}\cos x& \sin x\\ -\sin x& \cos x\end{array}\right|\\ & =& \left(\cos x\right)\times \left(\cos x\right)-\left(\sin x\right)\times \left(-\sin x\right)\\ & =& {\cos }^{2}x-\left(-{\sin }^{2}x\right)\\ & =& {\cos }^{2}x+{\sin }^{2}x\\ & =& 1\end{array}$$\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}\cos x& \sin x\\ -\sin x& \cos x\end{array}\right|\\ & =& \left(\cos x\right)\times \left(\cos x\right)-\left(\sin x\right)\times \left(-\sin x\right)\\ & =& {\cos }^{2}x-\left(-{\sin }^{2}x\right)\\ & =& {\cos }^{2}x+{\sin }^{2}x\\ & =& 1\\ & & \end{array}$

Finding the Wronskian from right side of given differential question

Find the Wronskian $W_1\text{and}{W}_2$of right side of the differential equation .$f\left(x\right)=\sec x$

$\begin{array}{rcl}{W}_1& =& \left|\begin{array}{cc}0& y_2\\ f\left(x\right)& y{\text{'}}_2\end{array}\right|\\ & =& \left|\begin{array}{cc}0& \sin x\\ secx& \cos x\end{array}\right|\\ & =& 0-\sin x\times \sec x\\ & =& -\sin x\times \frac{1}{\cos x}\\ & =& -\tan x\end{array}$

And

$\begin{array}{rcl}{W}_2& =& \left|\begin{array}{cc}{y}_1& 0\\ y{\text{'}}_1& f\left(x\right)\end{array}\right|\\ & =& \left|\begin{array}{cc}\cos x& 0\\ -\sin x& secx\end{array}\right|\\ & =& \cos x\times \sec x-0\\ & =& \cos x\times \frac{1}{\cos x}\\ & =& 1\end{array}$

Finding the value of u1 and  u2

Now, find the value of

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

And,

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

By integration, we get

$\begin{array}{rcl}{u}_1& =& \int u_1^{\text{'}}dx\\ & =& \int -\tan xdx\\ & =& \int \frac{-\sin x}{\cos x}dx\\ & =& \ln \left(\cos x\right)\\ & & \end{array}$

And,

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

Finding the required solution

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

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

Now, the general solution is:

$\begin{array}{rcl}y& =& y_c+y_p\\ & =& c_1\cos x+c_2\sin x+\cos x\ln \left(\cos x\right)+x\sin x\end{array}$