Author Topic: Q4-T0201/T0901  (Read 1463 times)

Victor Ivrii

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Q4-T0201/T0901
« on: March 02, 2018, 05:24:52 PM »
Find the general solution of the given differential equation.
$$y'' + 4y = 3 \csc (2t),\qquad 0< t < \pi /2.
$$
« Last Edit: March 02, 2018, 05:38:36 PM by Victor Ivrii »

Meng Wu

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Re: Q4-T0201
« Reply #1 on: March 02, 2018, 05:36:54 PM »
$$y''+4y=3\csc2t, 0< t<{\pi \over 2}$$
For homogeneous equation: $y''+4y=0$ $\\$
Characteristic equation: $$r^2+4=0 \implies \cases{r_1=2i\\r_2=-2i}$$
Complementary solution: $$\begin{align}y_c(t)&=c_1y_1(t)+c_2y_2(t)\\&=c_1\cos2t+c_2\sin2t\end{align}$$
For non-homogeneous equation: $y''+4y=3\csc2t$ $\\$
$p(t)=0, q(t)=4,g(t)=3\csc2t$ are continuous on $0< t<{\pi \over 2}$. $\\$
Now, $$W[y_1,y_2](t)=\begin{array}{|c c|} y_1(t) & y_2(t) \\ y_1'(t) & y_2'(t)\end{array}=\begin{array}{|c c|} \cos2t & \sin2t \\ -2\sin2t & 2\cos2t \end{array}=2$$
Thus, $y_1(t)$ and $y_2(t)$ form a fundamental set of solutions. $\\$
Therefore, $$\begin{align}u_1(t)=-\int{{y_2(t)g(t)\over W[y_1,y_2](t)}}&=-\int{{\sin2t\cdot 3\csc2t\over 2}}dt\\&=-\int{{\sin2t\cdot {3\over \sin2t}\over 2}}dt \\&=-\int{{3\over 2}}dt\\&=-{3\over2}t\end{align}$$
$$\begin{align}u_2(t)=\int{{y_1(t)g(t)\over W[y_1,y_2](t)}}&=\int{{\cos2t\cdot 3\csc2t\over 2}}dt\\&={3\over 2}\int{{\cos2t \over \sin2t}}dt \\&={3 \over 2}\int{\cot2t}dt\\&={3 \over 4}\ln|\sin2t|\end{align}$$
Hence, the particular solution is $y_p(t)=u_1(t)y_1(t)+u_2(t)y_2(t)$ $\\$
Thus, $$\begin{align}y_p(t)&=\cos2t\cdot (-{3\over 2}t)+\sin2t\cdot({3 \over 4}\ln|\sin2t|)\\&={3 \over 4}\sin2t\ln|\sin2t|-{3\over 2}t\cos2t\end{align}$$
Therefore, the general solution is:
$$\begin{align}y(t)&=y_c(t)+y_p(t)\\&=c_1\cos2t+c_2\sin2t+{3 \over 4}\sin2t\ln|\sin2t|-{3\over 2}t\cos2t\end{align}$$