# Toronto Math Forum

## MAT244-2018S => MAT244--Tests => Term Test 2 => Topic started by: Victor Ivrii on March 21, 2018, 03:03:17 PM

Title: TT2--P1M
Post by: Victor Ivrii on March 21, 2018, 03:03:17 PM
a. Find general solution of
$$y''+ 4y=8\cos^{-1}(t)\qquad -\frac{\pi}{2}<t<\frac{\pi}{2}.$$

b. Find solution, such that $y(0)=y'(0)=0$.
Title: Re: TT2--P1M
Post by: Meng Wu on March 21, 2018, 08:14:49 PM
\underline{\text{Solution:}}$$\\ Part(a) \\ First consider homogeneous equation:$$y''+4y=0$$characteristic equation:$$r^2+4=0 \implies \cases{r_1=2i\\r_2=-2i}$$Thus, the complementary solution$$y_c(t)=c_1\cos 2t+c_2\sin 2t$$Now consider the nonhomogeneous equation$$y''+y=8\cos^{-1}(t)$$Since y_1(t)=\cos 2t and y_2(t)=\sin 2t, Wronskain$$W=[y_1,y_2](t)=\begin{array}{|c c|}\cos 2t&\sin 2t\\-2\sin 2t&2\cos 2t\end{array}=2\cos^22t+2\sin^22t=2 \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(2t)\int_{t_0}^{t}{\sin(2s)\cdot {8\cos^{-1}(s)}\over 2}ds+\sin(2t)\int_{t_0}^{t}{\cos(2s)\cdot {8\cos^{-1}(s)}\over 2}ds\\&=-\cos(2t)\int_{t_0}^{t}{2\sin(s)\cos(s)\cdot {8\over \cos(s)}\over 2}ds+\sin(2t)\int_{t_0}^{t}{(2\cos^2(s)-1)\cdot {8\cos^{-1}(s)}\over 2}ds\\&=-8\cos(2t)\int_{t_0}^{t}\sin(s)ds+4\sin(2t)\int_{t_0}^{t}[2\cos(s)-\sec(s)]\\&=8\cos(2t)\cos(t)+4\sin(2t)[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(2t)+c_2\sin(2t)+8\cos(2t)\cos(t)+4\sin(2t)[2\sin(t)-\ln(\sec(t)+\tan(t))]\end{align}$$Part(b)\\$$\begin{align}y(0)&=c_1\cos(0)+c_2\sin(0)+8\cos(0)\cos(0)+4\sin(0)[2\sin(0)-\ln(\sec(0)+\tan(0))]\\&=c_1+8=0 \implies c_1=-8\end{align}y'(t)=-2c_1\sin(2t)+2c_2\cos(2t)-16\sin(2t)\cos(t)-8\cos(2t)\sin(t)+8\cos(2t)[2\sin(t)-\ln(\sec(t)+\tan(t)]+4\sin(2t)[2\cos(t)-{\sec(t)\tan(t)+\sec^2(t)\over \sec(t)+\tan(t)}]y'(0)=2c_2=0 \implies c_2=0$$Therefore, the general solution to the IVP is$$y(t)=-8\cos(2t)+8\cos(2t)\cos(t)+4\sin(2t)[2\sin(t)-\ln(\sec(t)+\tan(t))]\$