Author Topic: Problem 3 (afternoon)  (Read 1382 times)

Victor Ivrii

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Problem 3 (afternoon)
« on: October 23, 2019, 06:13:03 AM »
(a) Find the general solution for equation
\begin{equation*}
y'' -5y'+6 y= 52\cos (2x).
\end{equation*}
(b) Find solution, satisfying $y(0)=0$, $y'(0)=0$.
« Last Edit: October 23, 2019, 06:18:43 AM by Victor Ivrii »

Lan Cheng

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Re: Problem 3 (afternoon)
« Reply #1 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) $y’(x)=+2C_{1}e^{2x}+3C_{2}e^{3x}-2sin(2x)-10cos(2x).$

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

$C_{1}=-13,C_{2}=12.$

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

OK. V.I.
« Last Edit: October 31, 2019, 10:49:02 AM by Victor Ivrii »

Hongling Liu

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Re: Problem 3 (afternoon)
« Reply #2 on: October 23, 2019, 06:44:52 AM »
y’’ - 5y’ + 6y = 52cos2x
Solution:
a):
r^2 - 5r + 6 = 0
r1 = 3
r2 = 2
∴y(x) = C1•e^3x + C2•e^2x
∵ y’’ - 5y’ + 6y = 52cos2x
set Yp(x) = Acos2x + Bsin2x
Y’ = -2Asin2x + 2Bcos2x
Y’’ = -4Acos2x - 4Bsin2x
 -4Acos2x - 4Bsin2x -5(-2Asin2x + 2Bcos2x) + 6(Acos2x + Bsin2x) = 52cos2x
A = 1
B = -5
∴Y(x) = C1•e^3x + C2•e^2x + cos2x - 5sin2x
b):
When Y(0) = 0, Y’(0)= 0
∴C1 =12, C2= -13
∴Y(x) = 12•e^3x - 13•e^2x + cos2x - 5sin2x

Ruojing Chen

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Re: Problem 3 (afternoon)
« Reply #3 on: October 23, 2019, 06:45:49 AM »
(a)when $$y''-5y'+6y=0$$
$$r^2-5r+6=0$$
$$(r-2)(r-3)=0$$
$$r_1=-2,r_2=-3$$
$$\therefore y_c(x)=c_1e^{-2x}+c_2e^{-3x}$$

when $$y''-5y'+6y=52Cos(2x)$$
$$y_p(x)=ACos(2x)+BSin(2x)$$
$$y'=-2ASin(2x)+2BCos(2x)$$
$$y''=-4ACos(2x)-4BSin(2x)$$
$$-4ACos(2x)-4BSin(2x)+10ASin(2x)-10BCos(2x)+6ACos(2x)+6BSin(2x)=52Cos(2x)$$
$$Cos(2x)(-4A-10B+6A)=52Cos(2x)$$
$$Sin(2x)(-4B+10A+6B)=0$$
$$\therefore A-5B=26, B+5A=0$$
$$A=1,B=-5$$
$$\therefore y_p(x)=Cos(2x)-5Sin(2x)$$

$$y=y_c(x)+y_p(x)=c_1e^{-2x}+c_2e^{-3x}+Cos(2x)-5Sin(2x)$$

(b) y(0)=0,y'(0)=0
$$y=c_1e^0+c_2e^0+1=0$$
$$c_1+c_2=-1$$
$$y'=-2c_1e^{2x}-3c_2e^{3x}+2Sin(2x)-10Cos(2x)=2c_1e^0+3c_2e^0-10=0$$
$$-2c_1-3c_2=10$$
$$\therefore c_2=-8
c_1=7$$

$$\therefore y=7e^{-2x}-8e^{-3x}+Cos(2x)-5Sin(2x)$$
« Last Edit: October 23, 2019, 07:11:04 AM by rj127 »

huoyanro

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Re: Problem 3 (afternoon)
« Reply #4 on: October 23, 2019, 06:48:04 AM »
1.let y'' - 5y' +6y=0
r^2-5r+6=0
r1=-1 r2=-3
thus y1(x)=C1e^(-2x) + C2e^(-3x)
let y'' - 5y' +6y= 52cos(2x)
let y2(x)=A cos(2x)+Bsin(2x)
y2'=-2Asin(2x)+2B cos(2x)
y2''=-4A cos(2x)-4Bsin(2x)
-4A cos(2x)-4Bsin(2x)+10Asin(2x)-10B cos(2x)+6A cos(2x)+6Bsin(2x)=52cos(2x)
-4A-10B+6A=52
-4B+10A+6B=0
thus A=1, B=-5
y2(x)=cos(2x)-5sin(2x)
thus y(x)=C1e^(-2x)+C2e^(-3x)+cos(2x)-5sin(2x)

huoyanro

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Re: Problem 3 (afternoon)
« Reply #5 on: October 23, 2019, 06:53:29 AM »
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

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Re: Problem 3 (afternoon)
« Reply #6 on: October 23, 2019, 07:27:06 AM »
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 $$y(0)=0, y'(0)=0$$
$$c_1+c_2+1=0$$
$$2c_1+3c_2-10=0$$
$$c_2=12, c_1=-13$$
$$y=-13e^{2x}+12e^{3x}+\cos{2x}-5\sin{2x}$$
« Last Edit: October 23, 2019, 07:44:32 AM by Di Qiu »

yangyiq5

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Re: Problem 3 (afternoon)
« Reply #7 on: October 23, 2019, 07:30:41 AM »
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

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Re: Problem 3 (afternoon)
« Reply #8 on: October 23, 2019, 07:46:30 AM »
Sorry,but there is a mistake at the first step, where $$(r-2)(r-3)=0$$, so $$r=2, r=3$$
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, $y(x)=7e^{-2x}-8e^{-3x}+cos(2x)-5sin(2x).$

Yuefan Wang

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Re: Problem 3 (afternoon)
« Reply #9 on: October 23, 2019, 07:52:33 AM »
Question:

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

Linqian Shen

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Problem3 (afternoon)
« Reply #10 on: October 23, 2019, 08:00:14 AM »
Question:

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

Linqian Shen

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Re: Problem 3 (afternoon)
« Reply #11 on: October 23, 2019, 08:02:11 AM »
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

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Re: Problem 3 (afternoon)
« Reply #12 on: October 23, 2019, 08:04:06 AM »
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 $c_2 =\frac{1}{2}$ $c_1 = \frac{3}{2}$
$y(t) = \frac{3}{2} e^{3x} + \frac{1}{2} e^{-x} -2e^x -2e^{-x}x$
« Last Edit: October 23, 2019, 09:01:53 AM by Nan Yang »

Wang Jingyao

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Re: Problem3 (afternoon)
« Reply #13 on: October 23, 2019, 09:43:40 AM »
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)$\because~~y(0)=0$ , $y'(0)=0$

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

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

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

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

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

Wang Jingyao

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Re: Problem 3 (afternoon)
« Reply #14 on: October 23, 2019, 09:57:22 AM »
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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