MAT244--2018F > Quiz-4

Q4 TUT 03014

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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:

--- Quote from: Yunqi(Yuki) Huang on October 26, 2018, 05:59:28 PM ---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)||

--- End quote ---

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

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