TT2--P1D

[Victor Ivrii]

a. Find general solution of
$$
y''+ y=2\cos^{-2}(t)\qquad -\frac{\pi}{2}<t<\frac{\pi}{2}.$$

b. Find solution, such that $y(0)=y'(0)=0$.

[Meng Wu]

$\underline{\text{Solution:}}$$\\$
Part(a) $\\$
First consider homogeneous equation: $$y''+y=0$$
characteristic equation: $$r^2+1=0 \implies \cases{r_1=i\\r_2=-i}$$
Thus, the complementary solution $$y_c(t)=c_1\cos t+c_2\sin t$$
Now consider the nonhomogeneous equation $$y''+y=2\cos^{-2}(t)$$
Since $y_1(t)=\cos t$ and $y_2(t)=\sin t$, Wronskain $$W=[y_1,y_2](t)=\begin{array}{|c c|}\cos t&\sin t\\-\sin t&\cos t\end{array}=\cos^2t+\sin^2t=1 \neq0$$
Therefore, $y_1(t)$ and $y_2(t)$ form a fundamental set of solutions.$\\$
Use the Method of Variation of Parameters: $\\$
The particular solution $$\begin{align}Y(t)&=-y_1(t)\int_{t_0}^{t}{y_2(s)g(s)\over W[y_1,y_2](s)}ds+y_2(t)\int_{t_0}^{t}{y_1(s)g(s)\over W[y_1,y_2](s)}ds\\&=-\cos(t)\int_{t_0}^{t}\sin(s)\cdot {2\cos^{-2}(s)}ds+\sin(t)\int_{t_0}^{t}{\cos(s) \cdot 2\cos^{-2}(s)}ds\\&=-2\cos(t)\int_{t_0}^{t}{\sin(s)\over\cos^2(s)}ds+2\sin(t)\int_{t_0}^{t}{\cos(s)\over \cos^2(s)}ds\\&=-2\cos(t)\int_{t_0}^{t}\sec(s)\tan(s)ds+2\sin(t)\int_{t_0}^{t}\sec(s)ds\\&=-2\cos(t)[\sec(t)]+2\sin(t)[\ln(\sec(t)+\tan(t))]\\&=-2+2\sin(t)[\ln(\sec(t)+\tan(t))]\end{align}$$
Therefore, the general solution $$\begin{align}y(t)&=y_c(t)+Y(t)\\&=c_1\cos(t)+c_2\sin(t)+2\sin(t)[\ln(\sec(t)+\tan(t))]-2\end{align}$$
Part(b)$\\$
$$\begin{align}y(0)&=c_1\cos(0)+c_2\sin(0)+2\sin(0)[\ln(\sec(0)+\tan(0))]-2\\&=c_1-2=0 \implies c_1=2\end{align}$$
$$y'(t)=-c_1\sin(t)+c_2\cos(t)+2\cos(t)[\ln(\sec(t)+\tan(t))]+2\sin(t){\sec(t)\tan(t)+\sec^2{(t)}\over \sec(t)+\tan(t)}$$
$$y'(0)=-c_1\sin(0)+c_2\cos(0)+2\cos(0)[\ln(\sec(0)+\tan(0))]+2\sin(0){\sec(0)\tan(0)+\sec^2{(0)}\over \sec(0)+\tan(0)} \implies c_2=0$$
Therefore, the general solution to the IVP is $$y(t)=2\cos(t)+2\sin(t)[\ln(\sec(t)+\tan(t))]-2$$

[Victor Ivrii]

I integrated differently
\begin{align*}
C_1=&-2\int \sin(t)\cos^{-2}(t)\,dt =2\int \cos^{-2}(t)\,d\cos(t)= -2\cos^{-1}(t)\,dt+c_1,\\
C_2 =&2\int \cos^{-1}(t)\,dt= 2\int \cos^{-2}(t)\, d\sin(t)=2\int \frac{d\sin(t)}{1-\sin^2(t)}\\
=&\int \Bigl[ \frac{1}{1+\sin(t)}+\frac{1}{1-\sin(t)}\Bigr]\,d\sin(t)=\ln \bigl(\frac{1+\sin(t)}{1-\sin(t)}\bigr)+c_2
\end{align*}
so the answer looks different but is the same due to
$$\ln \bigl(\frac{1+\sin(t)}{1-\sin(t)}\bigr)=\ln \bigl(\frac{(1+\sin(t))^2}{\cos^2(t)}\bigr)=
2\ln \bigl(\frac{1+\sin(t)}{\cos(t)}\bigr)=2\ln \bigl(\tan(t)+\sec(t)\bigr).$$