Author Topic: Q4-T0101/T5101  (Read 1905 times)

Victor Ivrii

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Q4-T0101/T5101
« on: March 02, 2018, 05:24:14 PM »
Find the general solution of the given differential equation.
$$y'' + 9y = 9 \sec^2 (3t)\qquad 0< t < \pi /6;$$
« Last Edit: March 02, 2018, 05:37:24 PM by Victor Ivrii »

Meng Wu

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Re: Q4-T0101/T5101
« Reply #1 on: March 02, 2018, 05:38:02 PM »
$$y''+9y=9\sec^23t, 0<t<{\pi \over 6}$$
For homogeneous equation: $y''+9y=0$ $\\$
Characteristic equation: $$r^2+9=0 \implies \cases{r_1=3i\\r_2=-3i}$$
Complementary solution: $$\begin{align}y_c(t)&=c_1y_1(t)+c_2y_2(t)\\&=c_1\cos3t+c_2\sin3t\end{align}$$
For non-homogeneous equation: $y''+9y=9\sec^23t$ $\\$
$p(t)=0, q(t)=9,g(t)=9\sec^23t$ are continuous on $0< t<{\pi \over 6}$. $\\$
Now, $$W[y_1,y_2](t)=\begin{array}{|c c|} y_1(t) & y_2(t) \\ y_1'(t) & y_2'(t)\end{array}=\begin{array}{|c c|} \cos3t & \sin3t \\ -3\sin3t & 3\cos3t \end{array}=3$$
Thus, $y_1(t)$ and $y_2(t)$ form a fundamental set of solutions. $\\$
Therefore, $$\begin{align}u_1(t)=-\int{{y_2(t)g(t)\over W[y_1,y_2](t)}}&=-\int{{\sin3t\cdot 9\sec^23t\over 3}}dt\\&=-\int{3\sin3t{1\over \cos^23t}}dt \\&=-3\int{\sec3t\tan3t}dt\\&=-\sec3t\end{align}$$
$$\begin{align}u_2(t)=\int{{y_1(t)g(t)\over W[y_1,y_2](t)}}&=\int{{\cos3t\cdot 9\sec^23t\over 3}}dt\\&=\int{3\cos3t{1\over \cos^23t}}dt \\&=\int{3\sec3t}dt\\&=\ln|\sec3t+\tan3t|\end{align}$$
Hence, the particular solution is $y_p(t)=u_1(t)y_1(t)+u_2(t)y_2(t)$ $\\$
Thus, $$\begin{align}y_p(t)&=\cos3t\cdot (-\sec3t)+\sin3t\cdot(\ln|\sec3t+\tan3t|)\\&=\sin3t(\ln|\sec3t+\tan3t|)-1\end{align}$$
Therefore, the general solution is:
$$\begin{align}y(t)&=y_c(t)+y_p(t)\\&=c_1\cos3t+c_2\sin3t+\sin3t(\ln|\sec3t+\tan3t|)-1\end{align}$$