Problem 3 (afternoon)

MAT244--2019F > Term Test 1

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Linqian Shen:
Question:

find a general solution of $y^{\prime \prime}-5 y^{\prime}+6 y=52 \cos (2 x)$
and satisfy $y(0)=0, y^{\prime}(0)=0$
$$
\begin{array}{c}{y^{\prime \prime}-5 y^{\prime}+6 y=0} \\{(r-2)(r-3)=0}\end{array}
$$
$$
\begin{array}{l}{r_{1}=2, r_{2}=3} \\ {y_{c}(t)=c_{1} e^{2 t}+c_{2} e^{3 t}}\end{array}
$$
$$
\begin{array}{l}\\ {y_{p}(t)=A \cos (2 x)+B \sin (2 x)} \\ {y^{\prime} p(t)=-2 A \sin (2 x)+2 B \cos (2 x)} \\ {y^{\prime \prime} p(t)=-4 A \cos (2 x)-4B \sin (2 x)}\end{array}
$$
$$
\begin{array}{rl}{-4 A \cos (2 x)-4 B \sin (2 x)+10 A \sin (2 t)-10} & {B \cos (2 x)+6 A \cos (2 x)+6 B \sin (2 x)} \\ {} & {=52 \cos (2 x)}\end{array}
$$
$$
(-4 A-10 B+6 A) \cos (2 x)+(-4 B+10 A+6 B) \sin (2 x)=52 \cos (2 x)
$$
$$
\left\{\begin{array}{l}{2 A-10 B=52} \\ {2 B+10 A=0}\end{array} \quad \left\{\begin{array}{l}{A=1} \\ {B=-5}\end{array}\right.\right.
$$
$$
\begin{array}{l}{y_p(x)=\cos (2 x)-5 \sin (2 x)} \\ {y(x)=y_c(x)+y_p(x)=c_{1} e^{2 t}+c_{2} e^{3 t}+\cos (2 x)-5 \sin (2 x)}\\{ y^{\prime}(x)=2 c_1 e^{2 t}+3 c_{2} e^{3 t}-2 \sin (2 x)-10 \cos (2 x)} \\ { y(0)=0, y^{\prime}(0)=0}\\{\therefore\left\{\begin{array}{l}{c_{1}+c_{2}=-1} \\ {2 c_{1}+3 c_{2}=10}\end{array}\therefore \left\{\begin{array}{l}{c_{1}=-13} \\ {c_{2}=12}\end{array}\right.\right.}\\{\therefore y(x)=-13 e^{2 t}+12 e^{3 t}+\cos (2 x)-5 \sin (2 x)}
\end{array}
$$

Nan Yang:
Question: y'' - 2y' - 3y = 16coshx
y(0) = 0 y'(0) = 0
Solution
Since $coshy = \frac{e^x + e^{-x}}{2}$
Then $y'' - 2y' - 3y = 16\frac{e^x - e^{-x}}{2}$ = $8e^x + 8 e^{-x}$
$y'' - 2y' - 3y = 0$
$r^3 - 2r - 3 = 0$
$r = 3 , -1$
$y_c = c_1 e^{3x} + c_2 e^{-x}$

$y'' -2y' -3y = 8e^x$.
Let $Y_1= Ae^x$
$Y_1'= Ae^x$
$Y_1''= Ae^x$
$A -2A -3A = 8$
$A = -2$
$Y_1= -2e^x$

$y'' -2y' -3y = 8e^{-x}$.
Let $Y_1= Ae^{-x} \cdot x$
Then $Y_1'= Ae^{-x} - Ae^{-x} \cdot x$
Then $Y_1''= -Ae^{-x} - Ae^{-x} + Ae^{-x}x$
$-2A -2A = 8$ $A = -2$
$Y_2= -2e^{-x}x$

$Y = -2e^x -2e^{-x}x $

y(t) = $c_1 e^{3x} + c_2 e^{-x} -2e^x -2e^{-x}x$
since $y(0) = 0, y'(0) = 0$
$c_1 + c_2 -2 = 0$
$3c_1 -c_2 -2 -2= 0$
Then $c_2 =\frac{1}{2}$ $c_1 = \frac{3}{2}$
$y(t) = \frac{3}{2} e^{3x} + \frac{1}{2} e^{-x} -2e^x -2e^{-x}x$

Wang Jingyao:
a)$r^2-5r+6=0$

$(r-2)(r-3)=0$

$r_1=2,~r_2=3$

$\therefore~~y(x)=C_1e^{2x}+C_2e^{3x}$

$y'’-5y'+6y=52\cos(2x)$

$y_p(x)=A\cos(2x)+B\sin(2x)$

$y_p'(x)=-2A\sin(2x)+2B\cos(2x)$

$y_p''(x)=-4A\cos(2x)-4B\sin(2x)$

Plug in:

$-4A\cos(2x)-4B\sin(2x)-5(-2A\sin(2x)+2B\cos(2x))+6(A\cos(2x)+B\sin(2x))=52\cos(2x)$

$\therefore~~(2A-10B)\cos(2x)+(2B+10A)\sin(2x)=52\cos(2x)$

$\left\{\begin{array}{l}2A-10B=52\\2B+10A=0\end{array}\right.$

$\therefore~~\left\{\begin{array}{l}A=1\\B=-5\end{array}\right.$

$\therefore~~y_p(x)=\cos(2x)-5\sin(2x)$

$\therefore~~y(x)=C_1 e^{2x}+C_2 e^{3x}+\cos(2x)-5\sin(2x)$

b)$\because~~y(0)=0$ , $y'(0)=0$

$y=C_1 e^{2x}+C_2 e^{3x}+\cos(2x)-5\sin(2x)$

$y'(x)=2C_1 e^{2x}+3C_2 e^{3x}-2\sin(2x)-10\cos(2x)$

$\therefore~~\left\{\begin{array}{l}0=C_1+C_2+1\\0=2C_1+3C_2-10\end{array}\right.$

$\therefore~~\left\{\begin{array}{l}C_1=-13\\C_2=12\end{array}\right.$

$y(x)=-13e^{2x}+12e^{3x}+\cos(2x)-5\sin(2x)$

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