MAT244--2019F > Term Test 1

Problem 3 (afternoon)

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huoyanro:
2. y'(x)= -2C1e^(-2x)-3C2e^(-3x)-2sin(2x)-10cos(2x)
C1+C2+1=0
-2C1-3C2-10=0
C1=7,C2=-8
thus y(x)=7e^(-2x)-8e^(-3x)+cos(2x)-5sin(2x)

Di Qiu:
a)
$$(r-2)(r-3)=0$$
$$r=2, r=3$$
$$y_c = c_1e^{2x}+c_2e^{3x}$$
$$y_{p}=A\sin{2x}+B\cos{2x}$$
$$y'_{p}=2A\cos{2x}-2B\sin{2x}$$
$$y''_{p2}=-4A\sin{2x}-4B\cos{2x}$$
Therefore, $$-4A\sin{2x}-4B\cos{2x}-10A\cos{2x}+10B\sin{2x}+6A\sin{2x}+6B\cos{2x}=52\cos{2x}$$
$$A=-5, B=1$$
$$y_{p}=c_1e^{-x}+c_2e^{6x}$$
$$y=c_1e^{2x}+c_2e^{3x}+\cos{2x}-5\sin{2x}$$
b)
$$y'=2c_1e^{2x}+3c_2e^{3x}-2\sin{2x}-10\cos{2x}$$
substitutes $$y(0)=0, y'(0)=0$$
$$c_1+c_2+1=0$$
$$2c_1+3c_2-10=0$$
$$c_2=12, c_1=-13$$
$$y=-13e^{2x}+12e^{3x}+\cos{2x}-5\sin{2x}$$

yangyiq5:
Rewrite the question:
 y’’ -5y’ + 6y = 52 cos(2x)
1.   $$r^{2} – 5r +6 = 0$$
$$(r-2)(r-3) = 0$$
$$r_{1} = 2$$
$$r_{2} = 3$$
$$y_{c}(t) = C_{1}e^{2t} + C_{2}e^{3t}$$
2.   y’’ -5y’ + 6y = 52 cos(2x)
$$y_{p}(t) = A cos(2x) +B sin(2x)$$
$$y_{p}’(t) = -2Asin(2x) +2B cos(2x)$$
$$y_{p}’’(t) = -4Acos(2x) -4Bsin(2x)$$
$$(-4A-10B+6A)cos(2x)+(-4B+10A+6B)sin(2x) = (2A -10B) cos(2x)+(2B+10A)sin(2x)$$
$$2A-10B = 52$$
$$2B+10A = 0$$
$$A =1$$
$$B = -5$$
$$y_{p}(t) = cos(2x) – 5sin(2x)$$
$$y = C_{1}e^{2t} + C_{2}e^{3t}+ cos(2x) – 5sin(2x)$$
$$y’ = 2C_{1}e^{2t} +3C_{2}e^{3t}-10 cos(2x) – 2sin(2x)$$
$$when x=0,y = 0, y’=0$$
$$C_{1} = -13$$
$$C_{2} = 12$$
$$y = -13e^{2t} + 12e^{3t}+ cos(2x) – 5sin(2x)$$

Di Qiu:
Sorry,but there is a mistake at the first step, where $$(r-2)(r-3)=0$$, so $$r=2, r=3$$

--- Quote from: Lan Cheng on October 23, 2019, 06:37:47 AM ---a) 1. let $y"-5y'+6y=0.$

$r^{2}-5r+6=0.$

$\begin{cases}
r_{1}=-2 & r_{2}=-3.\end{cases}$

Thus, $y_{c}(x)=C_{1}e^{-2x}+C_{2}e^{-3x}.$

2.let $y"-5y'+6y=52cos(2x).$

let $y_{p}(x)=Acos(2x)+Bsin(2x). y'=-2Asin(2x)+2Bcos(2x),y"=-4Acos(2x)-4Bsin(2x).$

$-4Acos(2x)-4Bsin(2x)+10Asin(2x)-10Bcos(2x)+6Acos(2x)+6Bsin(2x)=52cos(2x).$

$\begin{cases}
-4A-10B+6A=52 & -4B+10A+6B=0\end{cases}.$

$A=1,B=-5. $

$y_{p}(x)=cos(2x)-5sin(2x).$

Therefore, $y(x)=C_{1}e^{-2x}+C_{2}e^{-3x}+cos(2x)-5sin(2x).$

b) $y’(x)=-2C_{1}e^{-2x}-3C_{2}e^{-3x}-2sin(2x)-10cos(2x).$

$\begin{cases}
C_{1}+C_{2}+1=0 & -2C_{1}-3C_{2}-10=0\end{cases}.$

$C_{1}=7,C_{2}=-8.$

Therefore, $y(x)=7e^{-2x}-8e^{-3x}+cos(2x)-5sin(2x).$

--- End quote ---

Yuefan Wang:
Question:

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

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