Author Topic: Problem 4 (main)  (Read 1048 times)

Victor Ivrii

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Problem 4 (main)
« on: October 23, 2019, 06:24:24 AM »
(a) Find the general solution for equation
\begin{equation*}
y'' -6y' +10y =2e^{3x} +39\cos(x) .
\end{equation*}
(b) Find solution, satisfying $y(0)=0$, $y'(0)=0$.

ChenYang Li

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Re: Problem 4 (main)
« Reply #1 on: October 23, 2019, 07:03:40 AM »
TT1 Question 4 (Main)
a).
Solve
\begin{equation}
y''-6y'+10y=2e^{3x}+39cosx
\end{equation}
First solve for
\begin{equation}
r^2-6r+10=0\\
\begin{split}
r & =\frac{-(-6)\pm \sqrt{(-6)^2-4(1)(10)}}{2(1)}\\
& = \frac{6\pm \sqrt{-4}}{2}\\
& = 3\pm i\\
\end{split}\\
y_c=c_1e^{3x}cosx+c_2e^{3x}sinx
\end{equation}
Then, we solve
\begin{equation}
y''-6y'+10y=2e^{3x}\\
y_{p1}=Ae^{3x} \qquad y_{p1}'=3Ae^{3x} \qquad y_{p1}''=9Ae^{3x}\\
9Ae^{3x}-6(3Ae^{3x})+10(Ae^{3x})=2e^{3x}\\
9A-18A+10A=2\\
A=2\\
y_{p1}=2e^{3x}
\end{equation}
Next, solve
\begin{equation}
y''-6y'+10y=39cosx\\
y_{p2}=Bcosx+Csinx \qquad y_{p2}'=-Bsinx+Ccosx \qquad y_{p2}''=-Bcosx-Csinx\\
(-Bcosx-Csinx)-6(-Bsinx+Ccosx)+10(Bcosx+Csinx)=39cosx\\
-Bcosx-6Ccosx+10Bcosx=39cosx\\
-Csinx+6Bsinx+10Csinx=0sinx\\
9B-6C=39\\
6B+9C=0\\
B=3 \qquad C=-2\\
y_{p2}=3cosx-2sinx
\end{equation}
Now, we add to the general form:
\begin{equation}
\begin{split}
y& =y_c+y_{p1}+y_{p2}\\
& =c_1e^{3x}cosx+c_2e^{3x}sinx+2e^{3x}+3cosx-2sinx
\end{split}\\
\end{equation}

b).
Input:  y(0)=0  y'(0)=0
Get the final solution
\begin{equation}
\begin{split}
y(0)& =c_1e^0cos0+c_2e^{0}sin0+2e^{0}+3cos0-2sin0\\
& =c_1(1)+c_2(0)+2(1)+3(1)-2(0)\\
& =c_1+5\\
& =0\\
\end{split}\\
c_1=-5\\
\quad \\
y'=3c_1e^{3x}cosx-c_1e^{3x}sinx+3c_2e^{3x}sinx+c_2e^{3x}cosx+6e^{3x}-3sinx-2cosx\\
\begin{split}
y'(0)& =3c_1e^{0}(1)-c_1e^{0}(0)+3c_2e^{0}(0)+c_2e^{0}(1)+6e^{0}-3(0)-2(1)\\
& =3c_1+c_2+6-2\\
& =3(-5)+c_2+4\\
& =c_2-11\\
& =0
\end{split}\\
c_2=11\\
\end{equation}
Now we know the two constant c, we plug them into the solution, get:
\begin{equation}
y=-5e^{3x}cosx+11e^{3x}sinx+2e^{3x}+3cosx-2sinx
\end{equation}
OK. V.I.
« Last Edit: October 31, 2019, 10:52:45 AM by Victor Ivrii »

jinshuyu

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Re: Problem 4 (main)
« Reply #2 on: October 23, 2019, 10:17:25 AM »
This is my answer for question4

Zhangxinbei

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Re: Problem 4 (main)
« Reply #3 on: October 23, 2019, 01:36:50 PM »
a)
First, solve the homogeneous system:
                   y’’-6y’+10y = 0
                    r^2-6r+10 = 0
                         R = 3+/-i
                   r1=3 + i  r2=3 - i. 
And
             Yc = C1e^(3x)cosx + C2e^(3x)sinx.

Now solve non homogeneous part:
                   y’’-6 y’+10 y = 2e^3x
LetYp = Ae^3x
               Then Yp’=3Ae^3x.   Yp’’ = 9Ae^3x
Plug in to the equation we have
                9Ae^3x-18Ae^3x + 10Ae^3x = 2e^3x
                               A = 2
So, Yp = 2e^3x
Now, we solve Y’’ - 6y’ +10y = 39cosx
                      Let Yp = Bcosx + Csinx
                     Then Yp’=-Bsinx + Ccosx   Yp’’ = -Bcosx - Csinx

Solve the linear equation:
                             9B-6C = 39
                              6B +9C = 0
                             B = 3, C = -2
Yp = 3cosx - 2sinx
Combining all we have:
Y= c1e^3xcosx + c2e^3xsinx + 2e^3x + 3cosx -2sinx

b)
As y(0) = 0 and y’(0) = 0,
Get the final solution
                           C1 = -5
                              C2 = 11
y = -5e^3xcosx + 11e^3xsinx + 2e^3x + 3cosx -2sinx

Yuying Chen

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Re: Problem 4 (main)
« Reply #4 on: October 23, 2019, 02:31:23 PM »
$\text{(a)}\\$
$r^2-6r+10=0\\$
$r=\frac{6\pm\sqrt{36-40}}{2}=\frac{6\pm2i}{2}=3\pm i\\$
$\text{Homogeneous Equation: $y_c(x)=c_1e^{3x}\cos x+c_2e^{3x}\sin x$}\\ \\$

$y^{\prime\prime}-6y^{\prime}+10y=2e^{3x}\\$
$Y_1(x)=Ae^{3x}\\$
$Y_1^{\prime}(x)=3Ae^{3x}\\$
$Y_1^{\prime\prime}(x)=9Ae^{3x}\\ $
$9Ae^{3x}-18Ae^{3x}+10Ae^{3x}=2e^{3x}\\$
$\qquad(9A-18A+10A)e^{3x}=2e^{3x}\\$
$\qquad\qquad\qquad\qquad\quad A=2\\$
$Y_1(x)=2e^{3x}\\ \\$

$y^{\prime\prime}-6y^{\prime}+10y=39\cos x\\$
$Y_2(x)=B\cos x+C\sin x\\$
$Y_2^{\prime}(x)=-B\sin x+C\cos x\\$
$Y_2^{\prime\prime}(x)=-B\cos x-C\sin x\\ $

$-B\cos x-C\sin x+6B\sin x-6C\cos x+10B\cos x+10C\sin x=39\cos x$
\begin{cases}
  (-B-6C+10B)\cos x=39\cos x\\\
  (-C+6B+10C)\sin x=0
   \end{cases}
\begin{cases}
  B=3\\
  C=-2
   \end{cases}
$Y_2(x)=3\cos x-2\sin x\\$
$\text{General Solution:}\\$
$y(t)=c_1e^{3x}\cos x+c_2e^{3x}\sin x+2e^{3x}+3\cos x-2\sin x\\$

$\text{(b)}\\$
$y^{\prime}=3c_1e^{3x}\cos x-c_1e^{3x}\sin x+3c_2e^{3x}\sin x+c_2e^{3x}\cos x+6e^{3x}-\sin x-2\cos x\\$
$y(0)=0\Rightarrow c_1+2+3=0\\$
$y^{\prime}(0)=0\Rightarrow 3c_1+c_2+6-2=0\\$
\begin{cases}
  c_1=-5\\
 c_2=11
   \end{cases}

$\text{Thus,}\\$
$y(t)=-5e^{3x}\cos x+11e^{3x}\sin x+2e^{3x}+3\cos x-2\sin x$

BJM

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Re: Problem 4 (main)
« Reply #5 on: October 23, 2019, 02:38:02 PM »
Here is the solution.