Solve by the method of variation of parameters: y''' + y' =…

Solve by the method of variation of parameters: $y''' + y' = \tan x \sec x$.

Step 1: Characteristic equation and fundamental system

Substituting $y=e^{rx}$ into the homogeneous equation $y'''+y'=0$ gives

\[ r^3+r=0 \implies r(r^2+1)=0 \implies r_1=0,\quad r_{2,3}=\pm i. \]

The fundamental system is

\[ y_1=1,\qquad y_2=\cos x,\qquad y_3=\sin x. \]

Step 2: Wronskian

\[ W = \begin{vmatrix} 1 & \cos x & \sin x \\ 0 & -\sin x & \cos x \\ 0 & -\cos x & -\sin x \end{vmatrix}. \]

Expanding along the first column, \[W = 1\cdot\begin{vmatrix} -\sin x & \cos x \\ -\cos x & -\sin x \end{vmatrix} = \sin^2x+\cos^2x = 1.\]

Step 3: Variation-of-parameters system

Write $y=u_1(x)\cdot1+u_2(x)\cos x+u_3(x)\sin x$; the standard system for $u_1',u_2',u_3'$ with $f(x)=\tan x\sec x$ is

\[ \begin{aligned} u_1' + u_2'\cos x + u_3'\sin x &= 0 && \text{(I)}\\ -u_2'\sin x + u_3'\cos x &= 0 && \text{(II)}\\ -u_2'\cos x - u_3'\sin x &= \tan x\sec x && \text{(III)} \end{aligned} \]

Step 4: Solve for $u_2'$

From (II), $u_3'=u_2'\tan x$. Substituting into (III): \[-u_2'\cos x - u_2'\tan x\sin x = \tan x\sec x \implies -u_2'\left(\cos x+\frac{\sin^2x}{\cos x}\right) = \frac{\sin x}{\cos^2x}.\]

Since $\cos x+\dfrac{\sin^2x}{\cos x}=\dfrac{\cos^2x+\sin^2x}{\cos x}=\dfrac{1}{\cos x}$, this reduces to

\[ -\frac{u_2'}{\cos x} = \frac{\sin x}{\cos^2x} \implies u_2' = -\tan x. \]

Step 5: Solve for $u_3'$ and $u_1'$

\[ u_3' = u_2'\tan x = -\tan^2x = 1-\sec^2x. \]

From (I): \[u_1' = -u_2'\cos x - u_3'\sin x = \tan x\cos x - (1-\sec^2x)\sin x = \sin x - \sin x + \sin x\sec^2x = \frac{\sin x}{\cos^2x} = \tan x\sec x.\]

Step 6: Integrate

\[ u_1=\int\tan x\sec x\,dx=\sec x,\qquad u_2=\int(-\tan x)\,dx=\ln|\cos x|,\qquad u_3=\int(1-\sec^2x)\,dx=x-\tan x. \]

Step 7: Assemble the particular solution

\[ y^{*} = u_1\cdot1+u_2\cos x+u_3\sin x = \sec x + \cos x\ln|\cos x| + (x-\tan x)\sin x. \]

Expanding the last term, $\sin x\tan x=\dfrac{\sin^2x}{\cos x}$, so

\[ \sec x - \frac{\sin^2x}{\cos x} = \frac{1-\sin^2x}{\cos x} = \frac{\cos^2x}{\cos x} = \cos x. \]

Hence $y^{*} = \cos x + \cos x\ln|\cos x| + x\sin x = \cos x\bigl(1+\ln|\cos x|\bigr) + x\sin x.$

Step 8: Absorb the redundant term

The standalone $\cos x$ term is itself a homogeneous solution ($=y_2$), so it may be dropped from $y^{*}$ without loss of generality – it is already accounted for by the arbitrary constant $C_2$ in the homogeneous part. The essential particular solution is

\[ y^{*} = \cos x\ln|\cos x| + x\sin x. \]

Step 9: General solution

The solution is recorded in the result below.