# Toronto Math Forum

## MAT244-2018S => MAT244--Tests => Term Test 2 => Topic started by: Victor Ivrii on March 21, 2018, 02:59:18 PM

Title: TT2--P1D
Post by: Victor Ivrii on March 21, 2018, 02:59:18 PM
a. Find general solution of
$$y''+ y=2\cos^{-2}(t)\qquad -\frac{\pi}{2}<t<\frac{\pi}{2}.$$

b. Find solution, such that $y(0)=y'(0)=0$.
Title: Re: TT2--P1D
Post by: Meng Wu on March 21, 2018, 05:14:19 PM
\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))]-2Title: Re: TT2--P1D Post by: Victor Ivrii on March 24, 2018, 10:43:40 AM 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).\$