# Toronto Math Forum

## MAT244--2020F => MAT244--Test & Quizzes => Test 1 => Topic started by: RunboZhang on October 23, 2020, 04:52:15 PM

Title: LEC0201-Retest-ALT-Y-Question2
Post by: RunboZhang on October 23, 2020, 04:52:15 PM
$\textbf {Problem:} \\\\$
$\text{(a) Find Wronskian } W(y_1, y_2)(x) \text{ of a fundamental set of solutions} y_1(x), y_2(x) \text{ for ODE} \\$
$\begin{gather} y'' + 3tan(x)y' + (1 + 3tan^{2}{x})y = 0 \end{gather}$
$\text{(b) Check that } y_1(x) = cos(x) \text{ is a solution and find another linearly independent solution.}\\\\$
$\text{(c) Write the general solution, and find solution such that} y(0) = 0 \text{ , } y'(0) = 12\\\\$

$\textbf{Solution: } \\\\$
$\text{(a):}\\\\$

\begin{gather} \begin{aligned} y'' + 3 tan(x) \cdot y' + (1 + 3tan^{2}(x))y = 0 \\\\ \implies p(x) = 3tan(x) \\\\ \end{aligned} \end{gather}

\begin{gather} \begin{aligned} w &= c \cdot e^{- \int_ (p(x)) \, dx} \\\\ &= c \cdot e^{- \int_ (3tan(x)) \, dx} \\\\ &= c \cdot e^{3ln|cos(x)| + c'} \\\\ &= c \cdot cos^{3}(x) + c \cdot e^{c'} \end{aligned} \end{gather}

$\text{Let c = 1, then we have: }$

\begin{gather} \begin{aligned} w = cos^{3}(x) \end{aligned} \end{gather}

$\text{(b):}\\\\$

\begin{gather} \begin{aligned} y_1 = cos(x) \ y_1' = - sin(x) \ y_1'' = - cos(x) \end{aligned} \end{gather}

$\text{Plug in the equation, we have: }$

\begin{gather} \begin{aligned} y'' + 3 tan(x) \cdot y' + (1 + 3tan^{2}(x))y &= -cos(x) + 3 \frac{sin(x)}{cos(x)} \cdot (- sin(x)) + (1 + 3 \frac{sin^{2}(x)}{cos^{2}(x)}) cos(x) \\\\ &= - cos(x) - 3\frac{sin^{2}(x)}{cos(x)} + cos(x) + 3 \frac{sin^{2}(x)}{cos(x)} \\\\ &= 0 \\\\ \end{aligned} \end{gather}

$\text{Thus we have proved} y_1(x) = cos(x) \text{ is a solution.} \\\\$

$\text{Now we solve for } y_2 \text{: } \\\\$

\begin{gather} \begin{aligned} W(x,y) &= cos^{3}(x) \\\\ &= cos(x) \cdot y_2' - (- sin(x)) \cdot y_2 \\\\ &= cos(x) \cdot y_2' + sin(x) \cdot y_2 \\\\ \end{aligned} \end{gather}

$\text{Now we solve for } y_2$  \text{ : }

\begin{gather} \begin{aligned} cos^{3}(x) &= cos(x) \cdot y_2' - (- sin(x)) \cdot y_2 \\\\ cos(x) &= \frac{y_2'}{cos(x)} + \frac{sin(x)}{cos^{2}(x)} y_2 \\\\ &= (\frac{y_2}{cos(x)})' \\\\ y_2 &= cos(x) \cdot \int(cos(x)) ,\ dx \\\\ &= cos(x) \cdot (sin(x) + c)\\\\ &= cos(x) \cdot sin(x) \\\\ &= c \cdot cos(x) \end{aligned} \end{gather}

$\text{(c):}\\\\$

$\text{We have: } y(0) = 0 \ \text{ and plug in our general form, we have: }$

\begin{gather} \begin{aligned} 0 &= c_1 + c_2 \cdot 0 \\\\ c_1 &= 0 \end{aligned} \end{gather}

$\text{Then our general form becomes: } y = c_2(cos(x) \cdot sin(x))$
$\text{Also, we have initial value: } y'(0) = 12 \ \text{ By plugging it in, we have: }$

\begin{gather} \begin{aligned} 12 &= c_2 \cdot (-sin0 \cdot sin0 + cos0 \cdot cos0) &= c_2 \end{aligned} \end{gather}

$\text{Therefore we got our final answer: } y = 12 \cdot sin(x) \cdot cos(x)$