Question

In Problems 1-6, use the method of variation of parameters to determine a particular solution to the given equation.

${\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{+}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{=}\mathit{t}\mathit{a}\mathit{n}\mathit{x}$

Short answer

The particular solution is$y_p=\ln \left|\sec x\right|-\sin x\ln |\sec x+\tan x|$

Step-by-step solution

Definition

Variation of parameters, also known as variation of constants, is a general method to solve inhomogeneous linear ordinary differential equations.

Find complementary solution

The given equation is:$y^{\text{'}\text{'}\text{'}}+y^{\text{'}}=\tan x$

The auxiliary equation is$D^3+D=0$

So,$D=\pm i,0$

So $\{1,\cos x,\sin x\}$fundamental set.

Calculate Wornkians

The value of wronkians is:

$$\begin{gathered} W\left[\begin{array}{ccc}1& \cos x& sinx\end{array}\right]=\left|\begin{array}{ccc}1& \cos x& \sin x\\ 0& -\sin x& \cos x\\ 0& -\cos x& -\sin x\end{array}\right| \\ =1 \end{gathered}$$

$$\begin{gathered} W_1=(-1{)}^{3-1}W\left[\begin{array}{cc}\cos x& \sin x\end{array}\right] \\ =\left|\begin{array}{cc}\cos x& \sin x\\ -\sin x& \cos x\end{array}\right| \\ =1 \\ {W}_2=(-1{)}^{3-2}W\left[\begin{array}{cc}1& \sin x\end{array}\right] \\ =-\cos x \\ {W}_3=(-1{)}^{3-3}W\left[\begin{array}{cc}1& \cos x\end{array}\right] \\ =-\sin x \end{gathered}$$

For particular solution

The particular solution is given by:

$$\begin{gathered} y_p=1\int \frac{1\left(\tan x\right)}{1}dx+\cos x\int \frac{-\cos x\left(\tan x\right)}{1}dx+\sin x\int \frac{-\sin x·\tan x}{1}dx \\ {y}_p=\ln \left|\sec x\right|+{\cos }^{2}x+\sin x{I}_1 \\ {I}_1=\int -\sin x·\tan xdx \\ =\sin x-\ln (\sec x+\tan x) \end{gathered}$$

$y_p=\ln \left|\sec x\right|+{\cos }^{2}x+{\sin }^{2}x-{\sin }^{2}x\ln (\sec x+\tan x)$

Since 1 is in fundamental set solution so$y_p=\ln \left|\sec x\right|-\sin x\ln |\sec x+\tan x|$