Author Topic: LEC0201-Retest-ALT-Y-Question2  (Read 315 times)

RunboZhang

  • Sr. Member
  • ****
  • Posts: 51
  • Karma: 0
    • View Profile
LEC0201-Retest-ALT-Y-Question2
« 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}
$

$\text{Therefore we got our final answer:   } y = 12 \cdot sin(x) \cdot cos(x)$
« Last Edit: October 27, 2020, 07:16:53 PM by RunboZhang »