Author Topic: Problem 3 (noon)  (Read 541 times)

Victor Ivrii

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Problem 3 (noon)
« on: October 23, 2019, 06:12:04 AM »
(a) Find the general solution for equation
\begin{equation*}
y'' -4y'+3 y= 96\sinh (x).
\end{equation*}
(b) Find solution, satisfying $y(0)=0$, $y'(0)=0$.
« Last Edit: October 23, 2019, 06:16:38 AM by Victor Ivrii »

Jiwen Bi

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Re: Problem 3 (noon)
« Reply #1 on: October 23, 2019, 06:45:21 AM »
$
{y}''-{4y}'+3y=96sinh(x)\\\\
(a). Homogeneous part:\\
r^2-4r+3=96sinh(x)\\\\
r^2-4r+3=96(\frac{e^{x}-e^{-x}}{2})=48e^{x}-48e^{x}\\\\
let\,r^2-4r+3=0\\
=(r-1)(r-3),r_{1}=1,r_{2}=3\\
y_{c}=C_{1}e^{x}+C_{2}e^{3x}\\
Next\,we\,solve \, {y}''+{4y}'+3y=48e^{x}\\
Since \,we \,already \,have \,Ae^{x} in \,p_{c} \,solution\\
let y_{p}=Axe^{x}
then\,{y}'=Axe^{x}+Ae^{x},\,{y}''=2Ae^{x}+Axe^{x}\\
{y}''+{4y}'+3y=2Ae^{x}+Axe^{x}+4(Axe^{x}+Ae^{x})+3(Axe^{x})\\
-2Ae^{x}=48e^{x}\\
-2A=48
A=-24,y_{p}=-24e^{x}\\
now\,let\,{y}''-{4y}'+3y=-48e^{-x}\\
let\, y_{p}=Be^{-x}\\
then\,{y}'=-Be^{-x},{y}''=Be^{-x}\\
{y}''-{4y}'+3y=Be^{-x}-4(-Be^{-x})+3(Be^{-x})\\
8Be^{x}=-48e^{x}\\
B=-6
y_{P}=-6e^{x}\\
So\, the \,general \,solution \,is\,y=y_{c}+y_{P}=C_{1}e^{x}+C_{2}e^{3x}-24xe^{x}-6e^{-x}\\\\
(b)y=C_{1}e^{x}+C_{2}e^{3x}-24xe^{x}-6e^{-x}\\
{y}'=C_{1}e^{x}+3C_{2}e^{3x}-24e^{x}-24xe^{x}+6e^{-x}\\
let\,y(0)=0\\C_{1}+C_{2}-6=0,C_{1}+C_{2}=6\\
let\,y(0)'=0
C_{1}+3C_{2}=18\\
C_{1}= 0,C{2}=6\\
so,\,y=6e^{3x}-24xe^{x}-6e^{-x}$

OK. V.I.









« Last Edit: October 31, 2019, 10:45:49 AM by Victor Ivrii »

Yue Sagawa

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Re: Problem 3 (noon)
« Reply #2 on: October 23, 2019, 06:49:12 AM »
Solution:
$$96 \sinh x = 96 \frac{e^{x}-e^{-x}}{2} = 48e^{x}-48e^{-x}$$
$$y" -4y'+3y = 48e^{x}-48e^{-x}$$
(a). Homogeneous part:
$$y"-4y'+3y=0$$
$$r^2-4r+3=0$$
$$(r-1)(r-3)=0$$
$$r_1 = 1, r_2=3$$
$$ y_0 = c_1e^{x}+c_2e^{3x}$$

Next we solve $y" -4y'+3y = 48e^{x}$
Since we already have $e^{x}$ in our solution, let $y_1 = Axe^{x}$
Then $$y_1'=Axe^{x}+Ae^{x}$$
$$y_1" = 2Ae^{x}+Axe^{x}$$
$$(2Ae^{x}+Axe^{x})-4(Axe^x+Ae^{x})+3Axe^{x} = 48e^{x}$$
$$(2A-4A)e^{x}=48e^{x}$$
$$-2A = 48$$
$$A=-24$$
$$y_1=-24xe^{x}$$

Next we solve $y" -4y'+3y = -48e^{-x}$
Let $y_2=Be^{-x}$
Then
$$y_2' = -Be^{-x}$$
$$y_2" = Be^{-x}$$
$$Be^{-x} +4Be^{-x}+3Be^{-x}=-48e^{-x}$$
$$8B = -48$$
$$B = -6 $$
$$y_2= -6e^{-x}$$
So the general solution is $y = y_0+y_1+y_2 = c_1e^{x}+c_2e^{3x} -24xe^{x}-6e^{-x}$

(b). $$y'=c_1e^{x}+3c_2e^{3x} -24e^{x}-24xe^x+6e^{-x}$$
$$y(0) = c_1+c_2-6 = 0 \Rightarrow c_1+c_2 = 6$$
$$y'(0) = c_1+3c_2-24+6 = 0 \Rightarrow c_1+3c_2=18$$
We get $$c_1 = 0, c_2=6$$
So $$y = 6e^{3x} -24xe^{x}-6e^{-x}$$
« Last Edit: October 23, 2019, 07:16:14 AM by yuetsai »