Author Topic: Problem 4 (noon)  (Read 660 times)

Victor Ivrii

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Problem 4 (noon)
« on: October 23, 2019, 06:26:03 AM »
(a) Find the general solution for equation
\begin{equation*}
y'' -8y' +25y =18e^{4x} +104\cos(3x) .
\end{equation*}
(b) Find solution, satisfying $y(0)=0$, $y'(0)=0$.

Yijin Qiang

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Re: Problem 4 (noon)
« Reply #1 on: October 23, 2019, 06:48:44 AM »
$y''-8y'+25y=18e^{4x}+104cos(3x)\\
First \,,solve \,the\, homogenous \,eqution:\\
r^{2}-8r+25=0\\
r^{2}-8r+16=-9\\
(r-4)^{2}=-9,r-4=9i\\
r_{1}=3i+4\\
r_{2}=-3i+4\\
solution \,to \,the\, homogenous\, eqution \,is\\
y_{c}(x)=C_{1}e^{4x}cos{3x}+C_{2}e^{4x}sin{3x}
now\,we\,are\,going\,to\,solve\,y_{p}(x)\\
y''-8y'+25y=18e^{4x}\\
y_{p}(x)=Ae^{4x}\\
y'=4Ae^{4x}\\
y'=16Ae^{4x}\,,then y''-8y'+25y=16Ae^{4x}-8(4Ae^{4x})+25(Ae^{4x})\\
A=2\\
Y_{P}(x)=2e^{4x}\\
and\,let's\,find\,y_{p_{2}}(x),\\
y''-8y'+25y=104cos(3x),\\
let \,y_{P}=Acos(3x)+Bsin(3x)\\
y'_{p}=-3Asin(3x)+3Bcos(3x)\\
y''_{p}=-9Acos(3x)-9Bsin(3x)\\
y''-8y'+25y=-9Acos(3x)-9Bsin(3x)-8(-3Asin(3x)+3Bcos(3x))+25(-3Asin(3x)+3Bcos(3x)\\
A=2,B=-3\,y_{P}=2cos(3x)-3sin(3x)
y=y_{c}+y_{p}=C_{1}e^{4x}cos{3x}+C_{2}e^{4x}sin{3x}+2e^{4x}+2cos(3x)-3sin(3x)\\
let\,y(0)=0,y'(0)=0,\\
C_{1}=-4,C_{2}=\frac{1}{3}\\
So\,,the\,final\,solution\,is\,-4e^{4x}cos{3x}+\frac{1}{3}e^{4x}sin{3x}+2e^{4x}+2cos(3x)-3sin(3x)$

Correctly found particular and general solution, (b) incorrect V.I.
« Last Edit: October 31, 2019, 11:20:21 AM by Victor Ivrii »

Klaire

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Re: Problem 4 (noon)
« Reply #2 on: October 23, 2019, 07:43:53 AM »
a,First solve the homogenous part:
\begin{aligned}
r^{2}-8r+25 &=0\\
r^{2}-8r+16&=-9\\
(r-4)^{2} &=9i^2\\
r_1 &=3i+4 \\r_2 &=-3i+4
\end{aligned}
So the solution to homogenous part is:
\begin{aligned}
y_c(x) =c_1e^{4x}cos(3x)+c_2e^{4x}sin(3x)
\end{aligned}
Next we solve
\begin{aligned}
y^{\prime\prime}-8y^{\prime}+25y &=18e^{4x}\\
\end{aligned}
Let
\begin{aligned}
y_p(x)&=Ae^{4x}\\
y_p^{\prime}(x) &= 4Ae^{4x}\\
y_p^{\prime\prime}(x) &= 16Ae^{4x}\\
                                 
\end{aligned}
Thus we can have
\begin{aligned}
16Ae^{4x}-32Ae^{4x}+25Ae^{4x} &=18e^{4x}\\
A &= 2\\
\end{aligned}
Therefore, we can have:
\begin{aligned}
y_p(x) &=2e^{4x}\\
\end{aligned}

\begin{aligned}
y^{\prime\prime}-8y^{\prime}+25y &=104Cos(3x)\\
y_p(x) &= Acos(3x)+Bsin(3x)\\
y_p^{\prime}(x) &= -3Asin(3x)+3Bcos(3x)\\
y_p^{\prime\prime}(x) &= -9Acos(3x)-9Bsin(3x)\\
\end{aligned}
Thus we can have
\begin{aligned}
-9Acos(3x)-9Bsin(3x)-8(-3Asin(3x)+3Bcos(3x))+25Acos(3x)+25Bsin(3x) &=104cos(3x)\\
(16A-24B)cos(3x)+(16B+24A)sin(3x) &= 104cos(3x)\\
A &= 2\\
B &=-3\\
y_p(x) &= 2cos(3x)-3sin(3x)
\end{aligned}
From the above, we get
\begin{aligned}
y &= y_c(x)+y_p(x)=c_1e^{4x}cos(3x)+c_2e^{4x}sin(3x)+2e^{4x}+2cos(3x)-3sin(3x)
\end{aligned}

b,
Since
\begin{aligned}
y(0) &= 0\\
y^{\prime}(0) &= 0\\
\end{aligned}
we can get
\begin{aligned}
c_1 &= -4\\
c_2 &= 1/3
\end{aligned}
So the solution is
\begin{aligned}
y &= -4e^{4x}cos(3x)+1/3e^{4x}sin(3x)+2e^{4x}+2cos(3x)-3sin(3x)
\end{aligned}





« Last Edit: October 23, 2019, 08:07:22 AM by Klaire »

Yiheng Bian

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Re: Problem 4 (noon)
« Reply #3 on: October 23, 2019, 07:58:21 AM »
Step 1:from question we get
$$
r^2-8r+25=0
$$
we solve the equation get
$$
r_1=4+3i,r_2=4-3i
$$
So we let
$$
y_c=c_1e^{4x}cos{3x}+c_2e^{4x}sin{3x}
$$
Step2 we solve
$$
 y''-8y'+25y=18e^{4x}
$$
let
$$
y_p=Ae^{4x}
$$
So we can know get
$$
y_p'=4Ae^{4x}, y_p''=16Ae^{4x}
$$
We take into original equation and simplify
$$
9Ae^{4x}=18e^{4x}
$$
So we get
$$
A=2
$$
$$
y_p=2e^{4x}
$$
Step3 we solve
$$
y''-8y'+25y=104cos{3x}
$$
let
$$
y_p=Asin{3x}+Bcos{3x}
$$
So
$$
y_p'=3Acos3x-3Bsin3x
$$
$$
y_p''=-9Asin3x-9Bcos3x
$$
we take into equation and simplify
So
$$
 6B+4A=0,2B-3A=13
$$
Therefore
$$
A=-3,B=2
$$
Therefore, general solution is
$$
y=c_1e^{4x}cos3x+c_2e^{4x}sin3x
+2e^{4x}-3sin3x+2cos3x
$$


Since
$$
 y(0)=0 , y'(0)=0}
$$
we take into general solution and get
$$
c_1=-4,c_2=1/3
$$
So
$$
y=-4e^{4x}cos3x+1/3e^{4x}sin3x
+2e^{4x}-3sin3x+2cos3x
$$
« Last Edit: October 31, 2019, 11:19:08 AM by Victor Ivrii »

Jiwen Bi

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Re: Problem 4 (noon)
« Reply #4 on: October 23, 2019, 09:28:40 PM »
$y''-8y'+25y=18e^{4x}+104cos(3x)\\
transfer\,it\,to\, homogenous \,eqution:\\
r^{2}-8r+25=0\\
Solve\,it\\
r^{2}-8r+16=-9\\
(r-4)^{2}=-9,r-4=9i\\
r_{1}=3i+4\\
r_{2}=-3i+4\\
y_{c}(x) solution:\\
y_{c}(x)=C_{1}e^{4x}cos{3x}+C_{2}e^{4x}sin{3x}\\
Now\,find\,first\,y_{p}(x)\\
y''-8y'+25y=18e^{4x}\\
y_{p}(x)=Ae^{4x}\\
y'=4Ae^{4x}\\
y'=16Ae^{4x}\,,then y''-8y'+25y=16Ae^{4x}-8(4Ae^{4x})+25(Ae^{4x})\\
A=2\\
Y_{P}(x)=2e^{4x}\\
find\,second\,y_{p}(x),\\
y''-8y'+25y=104cos(3x),\\
let \,y_{P}=Acos(3x)+Bsin(3x)\\
y'_{p}=-3Asin(3x)+3Bcos(3x)\\
y''_{p}=-9Acos(3x)-9Bsin(3x)\\
y''-8y'+25y=-9Acos(3x)-9Bsin(3x)-8(-3Asin(3x)+3Bcos(3x))+25(-3Asin(3x)+3Bcos(3x)\\
A=2,B=-3\,y_{P}=2cos(3x)-3sin(3x)\\
y=y_{c}+y_{p}=C_{1}e^{4x}cos{3x}+C_{2}e^{4x}sin{3x}+2e^{4x}+2cos(3x)-3sin(3x)\\
let\,y(0)=0,y'(0)=0,\\
C_{1}=-4,C_{2}=\frac{1}{3}\\
y(x)=-4e^{4x}cos{3x}+\frac{1}{3}e^{4x}sin{3x}+2e^{4x}+2cos(3x)-3sin(3x)
$