Q4 TUT 03014

[Victor Ivrii]

Find the general solution of the given differential equation.
$$
y'' + 9y = 9 \sec^2 (3t),\qquad 0< t < \frac{\pi}{6}.
$$

[Yunqi(Yuki) Huang]

For the left-hand side, we could write as $r^2+9=0$.Thus $r=3i$ or $r=-3i.
$
Now  $Y_{c}=C_{1}cos3t+C_{2}sin3t
$
For the Right-hand side, $Y_{1}(t)=cos3t$, $Y_{2}(t)+sin3t$, $g(t)=9sec^2(3t)$
 $$W(t)=\left[
 \begin{matrix}
   cos3t & sin3t \\
   -3sin3t& 3cos3t
  \end{matrix}
  \right] \tag{3}=3
$$
$$W_{1}(t)=\left[
 \begin{matrix}
   0 & sin3t \\
   1 & 3cos3t
  \end{matrix}
  \right] \tag{3}=-sin3t
$$
$$W_{2}(t)=\left[
 \begin{matrix}
   cos3t & 0 \\
   -3sin3t& 1
  \end{matrix}
  \right] \tag{3}=cos3t
$$
So, the particular solution is $$Y_{p}(t)=Y_{1}\int\frac{g(s)W_{1}(s)}{W(s)}\,ds+Y_{2}\int\frac{g(s)W_2(s)}{W(s)}\,ds=-1+\sin(3t)\ln |\sec(3t)+\tan(3t)|$$
Thus, the general solution is $Y(t)=C_{1}cos(3t)+C_{2}sin(3t)+sin(3t)ln$||sec(3t)+tan(3t)||-1

[Yunqi(Yuki) Huang]

For the left-hand side, we could write as $r^2+9=0$.Thus $r=3i$ or $r=-3i.
$
Now  $Y_{c}=C_{1}cos3t+C_{2}sin3t
$
For the Right-hand side, $Y_{1}(t)=cos3t$$.  Y_{2}(t)+sin3t$$. g(t)=9sec^2(3t)$
 $$W(t)=\left[
 \begin{matrix}
   cos3t & sin3t \\
   -3sin3t& 3cos3t
  \end{matrix}
  \right] \tag{3}=3
$$
$$W_{1}(t)=\left[
 \begin{matrix}
   0 & sin3t \\
   1 & 3cos3t
  \end{matrix}
  \right] \tag{3}=-sin3t
$$
$$W_{2}(t)=\left[
 \begin{matrix}
   cos3t & 0 \\
   -3sin3t& 1
  \end{matrix}
  \right] \tag{3}=cos3t
$$
So, the particular solution is $Y_{p}(t)=Y_{1}\int\frac{g(s)W_{1}(s)}{W(s)}\,ds+Y_{2}\int\frac{W(s)}\,ds=-1+sin(3t)ln$||sec(3t)+tan(3t)||
Thus, the general solution is $Y(t)=C_{1}cos(3t)+C_{2}sin(3t)+sin(3t)ln$|sec(3t)+tan(3t)|-1

[Victor Ivrii]

1) Need to escape \sin (x) , \ln (x) etc producing $\sin (x)$,...

2) Do not put dollar signs inside math formula! Only around it


Fix it

3) Where the constants from integration?

[Michael Poon]

So, the particular solution is $Y_{p}(t)=Y_{1}\int\frac{g(s)W_{1}(s)}{W(s)}\,ds+Y_{2}\int\frac{W(s)}\,ds=-1+sin(3t)ln$||sec(3t)+tan(3t)||

Small correction, likely due to typesetting error, but I think you mean the following:

So, the particular solution is $Y_{p}(t)=Y_{1}\int\frac{g(s)W_{1}(s)}{W(s)}\,ds+Y_{2}\int\frac{g(s)W_{2}(s)}{W(s)}\,ds=-1+sin(3t)ln|sec(3t)+tan(3t)|$

[Victor Ivrii]

Better but still not perfect

http://forum.math.toronto.edu/index.php?topic=1443.0

And what about constants (it should be a general solution)?

[Michael Poon]

Right, I think it should be:

$Y(t) = Y_c + Y_p$

$Y(t) = c_1 \cos(3t) + c_2 \sin(3t)+ \sin(3t) \ln| \sec(3t)+ \tan(3t)|-1 + c_3$, where $c_1, c_2, c_3 \in {\rm I\!R}$

Since $c_3$ can absorb the "-1", it becomes:

$Y(t) = c_1 \cos(3t) + c_2 \sin(3t)+ \sin(3t) \ln| \sec(3t)+ \tan(3t)| + c_3$, where $c_1, c_2, c_3 \in {\rm I\!R}$

[Victor Ivrii]

$c_3$? There are only 2 const, and $y=1$ is not a solution to homogeneous equation. And to make $\mathbb{R}$ (etc) you type \mathbb{R} !

[Yunqi(Yuki) Huang]

For the left-hand side, we could write as $r^2+9=0$.Thus $r=3i$ or $r=-3i.
$
Now  $Y_{c}=C_{1}cos3t+C_{2}sin3t
$
For the Right-hand side, $Y_{1}(t)=cos3t$$.  Y_{2}(t)+sin3t$$. g(t)=9sec^2(3t)$
 $$W(t)=\left[
 \begin{matrix}
   cos3t & sin3t \\
   -3sin3t& 3cos3t
  \end{matrix}
  \right] \tag{3}=3
$$
$$W_{1}(t)=\left[
 \begin{matrix}
   0 & sin3t \\
   1 & 3cos3t
  \end{matrix}
  \right] \tag{3}=-sin3t
$$
$$W_{2}(t)=\left[
 \begin{matrix}
   cos3t & 0 \\
   -3sin3t& 1
  \end{matrix}
  \right] \tag{3}=cos3t
$$
We could check the integral table that $\int\tan(t)\sec(t)dt=sec(t)$. $\int\sec(t)dt=ln|sec(t)+tan(t)|$
So, the particular solution is $Y_{p}(t)=Y_{1}\int\frac{g(s)W_{1}(s)}{W(s)}\,ds+Y_{2}\int\frac{g(s)W_{2}(s)}{W(s)}\,ds=-3\cos(3t)\int\tan(3s)\sec(3s)ds+3\sin(3t)\int\sec(3t)ds=-3\cos3t*\frac{1}{3}sec(3t)+3\sin(3t)ln|\sec3t+tan3t|*\frac{1}{3}$
Thus, the general solution is $Y(t)=C_{1}\cos(3t)+C_{2}\sin(3t)+\sin(3t)ln|\sec(3t)+\tan(3t)|-1$