MAT244--2019F > Term Test 1

Problem 3 (afternoon)

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Victor Ivrii:
(a) Find the general solution for equation
\begin{equation*}
y'' -5y'+6 y= 52\cos (2x).
\end{equation*}
(b) Find solution, satisfying $y(0)=0$, $y'(0)=0$.

Lan Cheng:
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.

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

$$\therefore y=7e^{-2x}-8e^{-3x}+Cos(2x)-5Sin(2x)$$

huoyanro:
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)

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