### Author Topic: Problem 3 (morning)  (Read 3388 times)

#### Victor Ivrii

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##### Problem 3 (morning)
« on: October 23, 2019, 06:11:02 AM »
(a) Find the general solution for equation
\begin{equation*}
y'' -6y'+8 y= 48\sinh (2x).
\end{equation*}
(b) Find solution, satisfying $y(0)=0$, $y'(0)=0$.

Hint: $\sinh(x)=\dfrac{e^x-e^{-x}}{2}$.
« Last Edit: October 23, 2019, 06:16:08 AM by Victor Ivrii »

#### Ruojing Chen

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##### Re: Problem 3 (morning)
« Reply #1 on: October 23, 2019, 06:36:33 AM »
As$$Sinh(2x)=\frac{e^{x}-e^{-x}}{2}$$
$$48Sinh(2x)=48(\frac{e^{2x}-e^{-2x}}{2})=24e^{2x}-24e^{-2x}$$
when $$y''-6y'+8y=0$$ $$r^2-6r+8=0$$
$$(r-4)(r-2)=0$$
$$r_1=4,r_2=2$$
$$\therefore y_c(x)=c_1e^{4x}+c_2e^{2x}$$
when $$y''-6y'+8=24e^{2x}$$
$$y_p(x)=Ae^{2x}x$$
$$y'=2Ae^{2x}x+Ae^{2x}$$
$$y''=4Ae^{2x}x+2Ae^{2x}+2Ae^{2x}=4Ae^{2x}x+4Ae^{2x}$$
$$4Ae^{2x}x+4Ae^{2x}-12Ae^{2x}x-6Ae^{2x}+8Ae^{2x}x=24e^{2x}$$
$$-2Ae^{2x}=24e^(2x)$$
$$A=-12$$
$$\therefore y_p(x)=-12e^{2x}x$$
when$$y''-6y'+8=-24e^{-2x}$$
$$y_p(x)=Be^{-2x}$$
$$y'=-2Be^{-2x}$$
$$y''=4Be^{-2x}$$
$$4Be^{-2x}+12Be^{-2x}+8Be^{-2x}=-24e^{-2x}$$
$$24Be^{-2x}=-24e^{-2x}$$
$$B=-1$$
$$\therefore y_p(x)=-e^{-2x}$$

$$y=y_c(x)+y_p(x)=c_1e^{4x}+c_2e^{2x}-12e^{2x}x-e^{-2x}$$

(b)when y(0)=0 $$c_1e^0+c_2e^0-12e^o*0-e^0=0$$
$$c_1+c_2=1$$
when y'(0)=0
$$y'=4c_1e^{4x}+2c_2e^{2x}-24e^{2x}x-12e^{2x}+2e^{-2x}$$
$$4c_1e^0+2c_2e^0+24e^0*0-12e^0+2e^0=0$$
$$4c_1+2c_2-12+2=0$$
$$2c_1+c_2=5$$
$$\therefore c_1=4,c_2=-3$$

$$y=4e^{4x}-3e^{2x}+12e^{2x}x-e^{-2x}$$ Some errors. V.I.
« Last Edit: October 31, 2019, 10:42:12 AM by Victor Ivrii »

#### Mengyuan Wang

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##### Re: Problem 3 (morning)
« Reply #2 on: October 23, 2019, 07:44:36 AM »

\begin{array}{l}{y^{\prime \prime}-6 y^{\prime}+8 y=48 \sinh (2 x)} \\ {y^{\prime \prime}-6 y^{\prime}+8 y=24 e^{2 x}-24 e^{-x}}\end{array}

\text { Let } y=e^{r x}

\begin{array}{l}{y^{\prime}=\sec ^{r x}} \\ {y^{\prime \prime}=r^{2} e^{r x}}\end{array}

\begin{array}{c}{y^{2}-6 x+8=0} \\ {r_{1}=2 \quad r_{2}=4} \\ {y=c_{1} e^{2x} +c_{2} e^{4x} }\end{array}

\text { let } y=A x e^{2 x}

y^{\prime}=A e^{2 x}+2 A x e^{2 x}

y^{\prime \prime}=4 A xx^{2 x}+4 A e^{2 x}

\begin{array}{c}{(8 A-12 A+4A) t \cdot e^{2 x}+(-6 A+4 A) e^{2 x}=24 e^{2 x}} \\ {-2 Ae^{2 x}=24 e^{2 x}} \\ {A=-12} \\ {y=-12 x e^{2 x}}
\end{array}

Let

\begin{array}{l}{y=A e^{-2 x}} \\ {y^{\prime}=-2 A e^{-2 x}} \\ {y^{\prime \prime}=4 A e^{-2 x}}\end{array}

\begin{aligned}(8 A+12 A+4 A) e^{-2 x} &=-24 e^{-2 x} \\ A &=-1 \\ y &=e^{-2 x} \end{aligned}

so

\begin{array}{l}{y=a e^{2}+c_{2} e^{4 x}-12 e^{2 x}-e^{-2 x}} \\ {y^{\prime}=2 c_{1} e^{2 t}+4 c_{2} e^{4 t}-24 r e^{2 x}+2 e^{-2 x}-12 e^{2 x}} \\ {y(0)=y^{\prime}(0)=0} \\ {y=-3 e^{2 x}+4 e^{4 x}-12 x e^{2 x}-e^{-2 x}}\end{array}

OK. V.I.
« Last Edit: October 31, 2019, 10:42:48 AM by Victor Ivrii »

#### Yiyang Huang

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##### Re: Problem 3 (morning)
« Reply #3 on: October 23, 2019, 08:05:12 AM »
Find the general solution for the equation
$$y^{\prime \prime}-6 y^{\prime}+8 y=48 \sinh (2 x)$$
$$\begin{array}{c}{r^{2}-6 r+8=0} \\ {(r-4)(r-2)=0} \\ {r=4 \quad r=2} \\ {y_c=c_{1} e^{4 x}+c_{2} e^{2 x}}\end{array}$$
$$y^{\prime \prime}-6 y^{\prime}+8 y=48 \sinh (2 x)$$
$$\begin{array}{l}{\sinh (x)=\frac{e^{x}-e^{-x}}{2}} \\ {\sinh (2 x)=\frac{e^{2 x}-e^{-2 x}}{2}} \\ {48 \sinh (2 x)=24 e^{2 x}-24 e^{-2 x}}\end{array}$$
$$y^{\prime \prime}-6 y^{\prime}+8 y=24 e^{2 x}$$

let $y_{p}(t)=A x e^{2 x}$
$$y^{\prime}=A e^{2 x}+2 A x e^{2 x} \quad y^{\prime \prime}=4 A e^{2 x}+4 A x e^{2 x}$$
\begin{aligned} 4 A e^{2 x}+4 A x e^{2 x}-6 A e^{2 x}-12 A x e^{2 x}+8 A x e^{2 x} &=24 e^{2 x} \\ 4 A-6 A &=24 \\ A &=-12 \\ y_{P}(t) &=-12 \times 2^{2 x} \end{aligned}
$$y^{\prime \prime}-6 y^{\prime}+8 y=-24 e^{-2 x}$$
Let $y_p(t)=A e^{-2 x} \quad y^{\prime}=-2 A e^{-2 x} \quad y^{\prime \prime}=4 A e^{-2 x}$
\begin{aligned} 44 e^{-2 x}+12 A e^{-2 x}+8 A e^{-2 x} &=-24 e^{-2 x} \\ A &=-1 \\ y_{p} &=-e^{-2 x} \end{aligned}
$$y=c_{1} e^{4 x}+c_{2} e^{2 x}-12 x e^{2 x}-e^{-2 x}$$

#### AllanLi

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##### Re: Problem 3 (morning)
« Reply #4 on: October 23, 2019, 09:20:39 AM »

y''-6y'+8y=48sinh(2x)
applying the hint

y''- 6y' + 8y= 24e^{2x}-24e^{-2x}
solving the homogeneous equation

r^2-6r+8=0 ,  r_1 = 2, r_2 = 4

Yc = C_1e^2 + C_2e^4
let Yp = Axe^{2x}

Yp' = A(e^{2x} + 2xe^{2x}), Yp'' = A(2e^2x + 2(e^{2x} +2xe^{2x}))
fill in the original equation

4Ae^{2x}+4Axe^{2x} - 6A(e^{2x}+2xe^{2x})+8Axe^{2x}=24e^{2x}
solve for A

A = -12
let

Ys = Be^{-2x}

Ys' = -2Be^{-2x} , Ys'' = 4Be^{-2x}

4Be^{-2x} -6(-2Be^{-2x}) + 8Be^{-2x} = -24e^{-2x}
then we have

24B = -24, B = -1
so we have general solution

y = c1e^{2x} + c2e^{4x} -12xe^{2x}-e^{-2x}
Part(b)

y(0)=0, y'(0)=0
we get

C_1+C_2-1 =0 , 2C_1 + 4C_2 -12+2 = 0
solve for C1 and C2

C_1 = -3 , C_2 = 4

y(x) = -3e^{2x}+4e^{4x}-12xe^{2x}-e^{-2x}

#### GuangyuDu

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##### Re: Problem 3 (morning)
« Reply #5 on: October 23, 2019, 10:02:25 AM »
Question 3:
$y''-6y'+8y=48\sin h (2x), y(0)=0, y'(0)=0.$

Solution:
$r^2-6r+8=0$
$r=4, r=2$
$y_c(x)=c_1e^{4t}+c_2e^{2t}$
$48\sin h (2x)=24\cdot e^{2x}-24\cdot e^{-2x}$
$y_p(x)=Axe^{2x}, y_p'=2Axe^{2x}+Ae^{2x},$
$y_p''(x)=4Axe^{2x}+2Ae^{2x}$
$y''-6y'+8y=-2Ae^{2x}=24e^{2x}$
$A=-12,y_p(x)=-12xe^{2x}$
$y_p(x)=Be^{-2x},y_p'(x)=-2Be^{-2x},y_p''(x)=4Be^{-2x}$
$y''-6y'+8y=24Be^{-2x}=-24e^{-2x}$
$B=-1,y_p(x)=-e^{-2x}$
$y=y_c+y_p=c_1e^{4t}+c_2e^{2t}-12xe^{2x}-e^{-2x}$

Plug in $y(0)=0$, $y'(0)=0$.

we have $c_1=4,c_2=3$
$y=4e^{4t}+3e^{2t}-12xe^{2x}-e^{-2x}$

#### Carrie

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##### Re: Problem 3 (morning)
« Reply #6 on: October 23, 2019, 11:12:34 AM »
$$y''-6y'+8y=48sinh(2x)$$
$$y(0)=0 \ y'(0)=0 \ sinhx=\frac{{e^x-e^{-x}}}{2}$$
$r^2-6r+8=0$\\
(r-2)(r-4)=0\\
r=2 \ r=4 \\
$y(x)=c_1e^{2x}+c_2e^{4x}$\\
$48sinh(2x)=48(\frac{e^x-e^{-x}}{2})=48(\frac{e^{2x}}{2}-\frac{e^{-2x}}{2})=24e^{2x}-24e^{-2x}$\\
$y"-6y'+8y=24e^{2x}$\\
$y_1(x)=Axe^{2x}$\\
$y_1'(x)=2Axe^{2x}+Ae^{2x}$\\
$y_{1}''(x)=4Axe^{2x}+4Ae^{2x}$\\
Substitute \  them\  back \ to \ the \ original \ equation\\
$4Axe^{2x}+4Ae^{2x}-12Axe^{2x}-6Ae^{2x}+8Axe^{2x}=24e^{2x}$\\
$-2Ae^{2x}=24Ae^{2x}$\\
A=-12\\
$y_1(x)=-12xe^{2x}$\\
$y''-6y'+8y=-24e^{-2x}$\\
$y_2(x)=Ae^{-2x}$\\
$y_2(x)=-2Ae^{-2x}$\\
$y_2(x)=4Ae^{-2x}$\\
Substitute \  them\  back \ to \ the \ original \ equation\\
$4Ae^{-2x}+12-2Ae^{-2x}+8Ae^{-2x}=-24Ae^{-2x}$\\
$24Ae^{-2x}=-24e^{-2x}$\\
A=-1\\
$y_2(x)=-e^{-2x}$\\
$y(x)=c_1e^{2x}+c_2e^{4x}-12xe^{2x}-e^{-2x}$\\
$y(0)=c_1+c_2-1=0$\\
$y'(x)=2c_1e^{2x}+4c_2e^{4x}-24xe^{2x}-12e^{2x}+2e^{-2x}$\\
$y'(0)=2c_1+4c_2-12+2=0$\\
$c_1=-3 \ c_2=4$\\
$y(x)=-3e^{2x}+4e^{4x}-12xe^{2x}-e^{-2x}$

#### Xuefeng Fan

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##### Re: Problem 3 (morning)
« Reply #7 on: October 23, 2019, 03:50:41 PM »
Hi everyone here is the solution of question 3:
y''−6y'+8y=48sinh(2x),y(0)=0,y'(0)=0.

Solution:
R^2−6R+8=0
therefore R=4,R=2
y=C1e^4t+c2e^2t

yp=Axe^2x,y'=2Ae^2x+Ae^2𝑥,
y''=4Axe^2𝑥+2Ae^2x
A=−12,Y=−12xe^2x
y=Be^−2x,
y'=−2Be^−2x,
y''=4Be^−2x
plug in
B=−1,yp=−e^−2x
Y=C1e^4t+C2e^2t−12Xe^x−e−2x

Plug in the value

we have C1=4,C2=-3

#### Xinyu Jing

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##### Re: Problem 3 (morning)
« Reply #8 on: October 24, 2019, 12:45:24 PM »
𝑦′′−6𝑦′+8𝑦=48sinh(2𝑥)

𝑦′′−6𝑦′+8𝑦=$24𝑒^{2𝑥}−24𝑒^{−2𝑥}$

Let $𝑦=𝑒^{𝑟𝑥(2)}$

𝑦′=sec(𝑟𝑥) 𝑦′′=$𝑟^{2}𝑒^{𝑟𝑥(3)}$

$r^{2}−r𝑥+8=0$

$𝑟_{1}=2, 𝑟_{2}=4$

$𝑦=𝑐_{1}𝑒^{4𝑥}+𝑐_{2}𝑒^{2x}$

let 𝑦=$𝐴𝑥𝑒^{2𝑥}$

𝑦′=$𝐴𝑒^{2𝑥}+2𝐴𝑥𝑒^{2𝑥}$

𝑦″=$4𝐴𝑒^{2𝑥}𝑥+2𝐴𝑒^{2𝑥}+2𝐴𝑒^{2𝑥}=4𝐴𝑒^{2𝑥}𝑥+4𝐴𝑒^{2𝑥}$

$(8𝐴−12𝐴+4𝐴)𝑡⋅𝑒^{2𝑥}+(−6𝐴+4𝐴)𝑒^{2𝑥}=24𝑒^{2𝑥}−2𝐴𝑒^{2𝑥}=24𝑒^{2𝑥}$

𝐴=−12 $𝑦=−12𝑥𝑒^{2𝑥}$

Let 𝑦=$B𝑒^{−2𝑥}$

𝑦′=$−2B𝑒^{−2𝑥}$

𝑦′′=$4B𝑒^{−2𝑥}$

$4𝐵𝑒^{−2𝑥}+12𝐵𝑒^{−2𝑥}+8𝐵𝑒^{−2𝑥}=−24𝑒^{−2𝑥}$

$24𝐵𝑒^{−2𝑥}=−24𝑒^{−2𝑥}$

𝐵=−1

∴$𝑦𝑝(𝑥)=−𝑒^{−2𝑥}$

so
$𝑦=𝑎𝑒^{2}+𝑐_{2}𝑒^{4𝑥}−12𝑒^{2𝑥}−𝑒^{−2𝑥}$

$𝑦′=2𝑐_{1}𝑒^{2𝑡}+4𝑐_{2}𝑒^{4𝑡}−24𝑟𝑒^{2𝑥}+2𝑒^{−2𝑥}−12𝑒^{2𝑥}$

𝑦(0)=𝑦′(0)=0

$𝑦=−3𝑒^{2𝑥}+4𝑒^{4𝑥}−12𝑥𝑒^{2𝑥}−𝑒^{−2𝑥}$
« Last Edit: October 24, 2019, 12:50:31 PM by Xinyu Jing »