Solve by variation of parameters y'' + y = (1)/(cos 2x √(cos…

Solve by variation of parameters $y'' + y = \frac{1}{\cos 2x\,\sqrt{\cos 2x}}$.

Step 1: Homogeneous equation

The characteristic equation $r^2+1=0$ gives $r=\pm i$, so the fundamental system is

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

Step 2: Wronskian

\[ W(y_1,y_2)=\begin{vmatrix}\cos x & \sin x\\-\sin x & \cos x\end{vmatrix}=\cos^2x+\sin^2x=1. \]

Step 3: Variation-of-parameters system

Write $y=u_1\cos x+u_2\sin x$ with $f(x)=(\cos 2x)^{-3/2}$. By Cramer's rule with $W=1$: \[u_1' = -\frac{y_2f}{W} = -\frac{\sin x}{(\cos 2x)^{3/2}}, \qquad u_2' = \frac{y_1f}{W} = \frac{\cos x}{(\cos 2x)^{3/2}}.\]

Step 4: Integrate for $u_1$

Write $\cos 2x = 2\cos^2x-1$ and substitute $t=\cos x$, $dt=-\sin x\,dx$: $u_1 = -\int\frac{\sin x\,dx}{(2\cos^2x-1)^{3/2}} = \int\frac{dt}{(2t^2-1)^{3/2}}.$

The standard antiderivative $\displaystyle\int\frac{dt}{(at^2+b)^{3/2}}=\frac{t}{b\sqrt{at^2+b}}$ is checked by differentiating: with $F(t)=t/(b\sqrt{at^2+b})$, \[F'(t) = \frac{\sqrt{at^2+b}-t\cdot\dfrac{at}{\sqrt{at^2+b}}}{b(at^2+b)} = \frac{(at^2+b)-at^2}{b(at^2+b)^{3/2}} = \frac{1}{(at^2+b)^{3/2}}.\ \checkmark\]

With $a=2$, $b=-1$: \[u_1 = \frac{t}{(-1)\sqrt{2t^2-1}}+C = -\frac{\cos x}{\sqrt{2\cos^2x-1}} = -\frac{\cos x}{\sqrt{\cos 2x}}.\]

Step 5: Integrate for $u_2$

Write $\cos 2x = 1-2\sin^2x$ and substitute $t=\sin x$, $dt=\cos x\,dx$: $u_2 = \int\frac{\cos x\,dx}{(1-2\sin^2x)^{3/2}} = \int\frac{dt}{(1-2t^2)^{3/2}}.$

Now $a=-2$, $b=1$, so by the same formula: \[u_2 = \frac{t}{1\cdot\sqrt{1-2t^2}}+C = \frac{\sin x}{\sqrt{1-2\sin^2x}} = \frac{\sin x}{\sqrt{\cos 2x}}.\]

Step 6: Particular solution

\[ y^* = u_1\cos x+u_2\sin x = -\frac{\cos^2x}{\sqrt{\cos 2x}}+\frac{\sin^2x}{\sqrt{\cos 2x}} = \frac{\sin^2x-\cos^2x}{\sqrt{\cos 2x}} = \frac{-\cos 2x}{\sqrt{\cos 2x}} = -\sqrt{\cos 2x}. \]

Step 7: Verify $y^*$

With $y^*=-(\cos2x)^{1/2}$: $y^{*\prime} = -\tfrac12(\cos2x)^{-1/2}(-2\sin2x) = \frac{\sin2x}{\sqrt{\cos2x}}.$

Differentiating again (quotient rule) and using $\sin^22x=1-\cos^22x$: \[y^{*\prime\prime} = \frac{2\cos2x\sqrt{\cos2x}+\dfrac{\sin^22x}{\sqrt{\cos2x}}}{\cos2x} = \frac{2\cos^22x+\sin^22x}{(\cos2x)^{3/2}} = \frac{\cos^22x+1}{(\cos2x)^{3/2}}.\]

Hence \[y^{*\prime\prime}+y^* = \frac{\cos^22x+1}{(\cos2x)^{3/2}} - \sqrt{\cos2x} = \frac{\cos^22x+1-\cos2x\cdot\cos2x}{(\cos2x)^{3/2}} = \frac{1}{(\cos2x)^{3/2}} = \frac{1}{\cos2x\sqrt{\cos2x}}.\ \checkmark\]

Step 8: General solution

The solution is recorded in the result below.