Author Topic: LEC5201 QUIZ5  (Read 686 times)

Kun Zheng

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LEC5201 QUIZ5
« on: November 01, 2019, 02:03:36 PM »
Good afternoon,
Please enjoy the last day before reading week.
The question I had on lecture was:
Find the general solution of the given differential equation.
$y''+9y=9sec^23t,  0<t<\pi/6$
For the homogeneous side y''+9y=0:
$r^2+9=0$
$r^2=-9$
$r_1=3i, r_2=-3i$
so the complementary solution is
$y_c(t)=Acos3t+Bsin3t$
For the non-homogeneous side:
$p(t)=0, q(t)=9, g(t)=9sec^23t$ are continuous on $0<t<\pi/6$
$W_{y1,y2}(t)=
\begin{vmatrix}
y_1(t) & y_2(t) \\
y'_1(t) & y'_2(t)
\end{vmatrix}=
\begin{vmatrix}
cos3t & sin3t \\
-3sin3t & 3cos3t
\end{vmatrix}=3$
Thus, $W_1(t)=-\int \frac{y_2(t)g(t)}{W_{1,2}(t)}dt=-\int \frac{sin3t9sec^23t}{3}dt=-3\int \frac{sin3t}{cos^23t}dt$
$=-3\int tan3tsec3tdt=-sec3t$
And, $W_2(t)=\int \frac{y_1(t)g(t)}{W_{1,2}(t)}dt=\int \frac{3cos3t/cos^23t}{3}dt$
$=\int 3sec3t dt=ln|sec3t+tan3t|$
(Be careful the integral of trigonometric here, honestly I made a mistake may lose some mark here on the lecture quiz, we must memorize them)
In conclusion, $Y(t)=y_c(t)+y_p(t)$
$Y(t)=Acos3t+Bsin3t+cos3t(-sec3t)+ln|sec3t+tan3t|sin3t$
$Y(t)=Acos3t+Bsin3t+ln|sec3t+tan3t|sin3t-1$
This is all, hope you have a good reading week!