Question

Verify that the given two-parameter family of functions is the general solution of the non homogeneous differential equation on the indicated interval.

$$\begin{gathered} \mathbf{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{y}\mathbf{=}\mathbf{sec}\mathbf{}\mathbf{x} \\ \mathbf{y}\mathbf{=}{\mathbf{c}}_{\mathbf{1}}\mathbf{cos}\mathbf{}\mathbf{x}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}\mathbf{sin}\mathbf{}\mathbf{x}\mathbf{+}\mathbf{x}\mathbf{}\mathbf{sin}\mathbf{}\mathbf{x}\mathbf{+}\mathbf{(}\mathbf{cos}\mathbf{}\mathbf{x}\mathbf{)}\mathbf{}\mathbf{ln}\mathbf{(}\mathbf{cos}\mathbf{}\mathbf{x}\mathbf{)}\mathbf{,}\mathbf{}\mathbf{(}\mathbf{-}\mathbf{\pi }\mathbf{/}\mathbf{2}\mathbf{,}\mathbf{\pi }\mathbf{/}\mathbf{2}\mathbf{)} \end{gathered}$$

Short answer

Proved

Step-by-step solution

Deriving the given solution by differential equation

We have the non- homogeneous second order differential equation

$y\text{'}\text{'}+y=\sec x$

With an alleged general solution

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

Where the homogeneous part is$c_1\cos x+c_2\sin x$ and the particular part is

$x\sin x+\cos x\ln (\cos x)$

We have to make sure that it is the general solution of the given differential equation on the interval $\left(-\frac{\pi }{2},\frac{\pi }{2}\right)$as the following technique:

First, we have to verify that the functions of the homogeneous part $y_1=\cos xand{y}_2=\sin x$ satisfy the homogeneous differential equation$y\text{'}\text{'}+y=0$ as the following:

For the first function$y_1=\cos x$ .

We have to differentiate this function with respect to x , and then we have

$y_1\text{'}=-\sin x,y_1\text{'}\text{'}=-\cos x$

After that, we have to substitute with these derivatives into the given differential equation, and then we have

$\begin{array}{rcl}-\cos x+\cos x& =& 0\\ 0& =& 0\end{array}$

For the second function:$y_2=\sin x$

We have to differentiate this function with respect to x, then we have

$y_2\text{'}=\cos x,y_2\text{'}\text{'}=-\sin x$

After that, we have to substitute with these derivatives into the given differential equation, and then we have

$\begin{array}{rcl}-\sin x+\sin x& =& 0\\ 0& =& 0\end{array}$

Then the two functions of the homogeneous part satisfy the homogeneous D.E $y\text{'}\text{'}+y=0$.

Find the wronskian and solve the homogeneous differential equation

Second, we have to know if the functions of the homogeneous solution are linearly dependent or independent using the wronskian as

$$\begin{gathered} W\left(y_1,y_2\right)=\left|\begin{array}{cc}{y}_1& y_2\\ y_1\text{'}& y_2\text{'}\text{'}\end{array}\right| \\ =\left|\begin{array}{cc}\cos x& \sin x\\ -\sin x& \cos x\end{array}\right| \end{gathered}$$

After that, we have to find the determinate of this matrix as

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

Since the determinate of the wronskian of the given set of functions is not equal Zero, then this set of function is linearly independent, and the given functions can form a fundamental set of solution.

Finally, we have to verify that the particular solutions$y_p=x\sin x+\cos x\ln (\cos x)$ satisfy the given non-homogeneous differential equation $y\text{'}\text{'}+y=\sec x$ as the following:

Differentiate the particular solution with respect to x, and then we have

$\begin{array}{rcl}y{\text{'}}_p& =& \sin x+x\cos x-\sin x\ln (\cos x)+\cos x\times \frac{-\sin x}{\cos x}\\ & =& \sin x+x\cos x-\sin x\ln (\cos x)-\sin x\\ & =& x\cos x-\sin x\ln (\cos x)\end{array}$

$\begin{array}{rcl}{y}_p\text{'}\text{'}& =& \cos x-x\sin x-\cos x\ln (\cos x)-\sin x\times \frac{-\sin x}{\cos x}\\ & =& \cos x-x\sin x-\cos x\ln (\cos x)+\frac{{\sin }^{2}x}{\cos x}\end{array}$

 Step 3: Deriving the differential equation using the given intervals

After that, we have to substitute with $y_p=x\sin x+\cos x\ln (\cos x)andy\text{'}\text{'}p$ into the given differential equation as

$y\text{'}\text{'}+y=\sec x$

$\begin{array}{rcl}\left(\cos x-x\sin x-\cos x\ln (\cos x)+\frac{{\sin }^{2}x}{\cos x}\right)+(x\sin x+\cos x\ln (\cos x\left)\right)& =& \sec x\\ \cos x+\frac{{\sin }^{2}x}{\cos x}& =& \sec x\\ \frac{\cos x\times \cos x+{\sin }^{2}x}{\cos x}& =& \sec x\\ \frac{{\cos }^{2}x+{\sin }^{2}x}{\cos x}& =& \sec x\\ \frac{1}{\cos x}& =& \sec x\\ \sec x& =& \sec x\end{array}$

Then the particular parts$y_p=x\sin x+\cos x\ln (\cos x)$satisfy the given differential equation.

In conclusion, we find that the general solution $y=c_1\cos x+c_2\sin x+x\sin x+\cos x\ln (\cos x)$ is valid on the indicated interval$(0,\infty )$ .

Hence it is verified that the given two-parameter family of functions is the general solution of the non homogeneous differential equation on the indicated interval.