Question

Solve each differential equation by variation of parameters

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

Short answer

$y=c_1\cos x+c_2\sin x+\sin x\ln (\sec x+\tan x)-1$

Step-by-step solution

Step 1: Determine the corresponding homogeneous equation

We have the second order differential equation

$y^{\text{'}\text{'}}+y={\sec }^{2}x,$

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

$m^2+1=0$

Then the roots are

$m_{1,2}=\pm i$

which are conjugate and complex.

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

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

$y_h=c_1\cos x+c_2\sin x$ …….. (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 $y_1=\cos x$and $y_2=\sin x$, then we can obtain the wronskian as

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

After that, since we have the right side of the given differential equation$f\left(x\right)={\sec }^{2}x$ , then we can obtain the wronskian $W_1$and $W_2$as

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

and

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

After we had obtained the wronskian$W_1$ and$W_2$ , we have to obtain$u_1^{\text{'}}$ and $u_2^{\text{'}}$as the following

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

and

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

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

$\begin{array}{rcl}{u}_1& =& \int u_1^{\text{'}}dx\\ & =& \int -\sec x\tan xdx\\ & =& \int -\frac{1}{\cos x}\times \frac{\sin x}{\cos x}dx\\ & =& \int -\frac{\sin x}{{\cos }^{2}x}dx\end{array}$

Now using integration by substitution if we set $v=\cos x$then we have$dv=-\sin x$ , then we obtain

$\begin{array}{rcl}{u}_1& =& \int \frac{dv}{v^2}\\ & =& -\frac{1}{v}\\ & =& -\frac{1}{\cos x}\\ & =& -\sec x\end{array}$

Now we can obtain $u_2$as

$\begin{array}{rcl}{u}_2& =& \int u_2^{\text{'}}dx\\ & =& \int \sec xdx\end{array}$

We have to find the integration $\int \sec xdx$using integration by substitution as the following

$\begin{array}{rcl}\int \sec xdx& =& \int \sec x\times \frac{(\sec x+\tan x)}{(\sec x+\tan x)}\\ & =& \int \frac{\left({\sec }^{2}x+\sec x\tan x\right)}{(\sec x+\tan x)}\end{array}$

If we set$v=(\sec x+\tan x)$ , then we find that $dv=\left(\sec x\tan x+{\sec }^{2}x\right)$, then we have

$\begin{array}{rcl}\int \sec xdx& =& \int \frac{dv}{v}\\ & =& \ln \left(u\right)\\ & =& \ln (\sec x+\tan x)\end{array}$

Then we have

$u_2=\ln (\sec x+\tan x)$

Determine the general solution for given differential equation

After that, using the values of$u_1$ and $u_2$the functions role="math" localid="1664176732305" width="17" height="24">y1and role="math" localid="1664176744614" $y_2$ , we can obtain the particular solution as

$\begin{array}{rcl}{y}_p& =& u_1{y}_1+u_2{y}_2\\ & =& (-\sec x)\times \cos x+\ln (\sec x+\tan x)\times \sin x\\ & =& -\frac{1}{\cos x}\times \cos x+\sin x\ln (\sec x+\tan x)\\ & =& \sin x\ln (\sec x+\tan x)-1\end{array}$

$y_p=\sin x\ln (\sec x+\tan x)-1$ ……. (b)

Finally, from equations (a) and (b) we can obtain the general solution of the differential equation as$y^{\text{'}\text{'}}+y={\sec }^{2}x$

$\begin{array}{rcl}y& =& y_h+y_p\\ & =& c_1\cos x+c_2\sin x+\sin x\ln (\sec x+\tan x)-1\\ & & \end{array}$