# Toronto Math Forum

## MAT244--2020F => MAT244--Test & Quizzes => Test 2 => Topic started by: RunboZhang on November 19, 2020, 01:12:05 PM

Title: TT3-LEC0201-ALT-E-Q2
Post by: RunboZhang on November 19, 2020, 01:12:05 PM
$\textbf{Problem2:} \\\\ \text{Find the general solution of }$
$$y'''+y' = \frac{2}{sin(2t)}$$
$\text{Hint: Use method of variations.} \int{sec(t)}\,dt = ln(tan(t)+sec(t)) \text{ and } \int{csc(t)}\,dt = -ln(cot(t)+csc(t))$

$\textbf{Solution: }$

$\text{Consider the characteristic polynomial:}$
$$\lambda^3 + \lambda = 0 \\\\ \lambda(\lambda^2+1) = 0 \\\\ \lambda_1 =0, \lambda_2=i, \lambda_3=-i$$
$\text{Hence we have }$
$$y_1 = c_1e^0 = c_1 \\\\ y_2 = c_2cos(t) \\\\ y_3 = c_3sin(t)$$

$\text{Take} c_1 = c_2 = c_3 = 1 \text{, we have}$
$$w_1 =\begin{bmatrix} cos(t) & sin(t) \\ -sin(t) & cos(t) \\ \end{bmatrix} = 1$$

$$w_2 = -\begin{bmatrix} 1 & sin(t) \\ 0 & cos(t) \\ \end{bmatrix} = -cos(t)$$

$$w_3 = \begin{bmatrix} 1 & cos(t) \\ 0 & -sin(t) \\ \end{bmatrix} = -sin(t)$$

$\text{Then we have Wronskian of }y_1 ,\ y_2 ,\ y_3$
\begin{gather} \begin{aligned} W&= y_1''w_1 + y_2''w_2+y_3''w_3 \\\\ &= 0+ (cos(t))''(-cos(t)) + (sin(t))''(-sin(t)) \\\\ &= cos^2(t)+sin^2(t) \\\\ & =1 \end{aligned} \end{gather}

$\text{To compute }u_1 ,\ u_2,\ u_3 \text{, we need to use the hind in integral computation}$

\begin{gather} \begin{aligned} u_1 &= \int{\frac{1\cdot \frac{2}{sin(2t)}}{1}} \, dt \\\\ &= 2 \int{csc(2t)} \,dt \ \ \ \text{ (by the hint)} \\\\ &= -2 \cdot ln(cot(2t)+csc(2t)) + c_1 \end{aligned} \end{gather}

\begin{gather} \begin{aligned} u_2 &=\int{\frac{(-cos(t))\cdot \frac{2}{sin(2t)}}{1}} \, dt \\\\ & = -2\int{\frac{cos(t)}{sin(2t)}} \,dt \\\\ &= -2\int{\frac{cos(t)}{2sin(t)cos(t)}} \,dt \\\\ &= - \int{csc(t)} \,dt \ \ \ \text{ (by the hint)} \\\\ &= ln(cot(t) + csc(t)) + c_2 \end{aligned} \end{gather}

\begin{gather} \begin{aligned} u_3 &=\int{\frac{(-sin(t))\cdot \frac{2}{sin(2t)}}{1}} \, dt \\\\ & = -2\int{\frac{sin(t)}{2sin(t)cos(t)}} \,dt \\\\ & = - \int{sec(t)} \,dt \ \ \ \text{ (by the hint)} \\\\ &= -ln(tan(t) + sec(t)) +c_3 \end{aligned} \end{gather}

$\text{Therefore, the general solution is}$

\begin{gather} \begin{aligned} y &= u_1y_1 + u_2y_2 + u_3y_3 \\\\ &= -2 \cdot ln(cot(2t)+csc(2t)) + c_1 + cos(t)\cdot(ln(cot(t) + csc(t)) + c_2) + sin(t) \cdot (-ln(tan(t) + sec(t)) +c_3) \ \ \ \ \text{(by rearranging, we have)}\\\\ & =c_1 + c_2cos(t) + c_3sin(t) - 2 \cdot ln(cot(2t)+csc(2t)) + cos(t) \cdot ln(cot(t) + csc(t)) - sin(t) \cdot ln(tan(t) + sec(t) \end{aligned} \end{gather}