Author Topic: TT2B-P2  (Read 791 times)

Victor Ivrii

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TT2B-P2
« on: November 20, 2018, 05:55:03 AM »
Consider equation
\begin{equation}
y'''-y''  +4y'-4y= 8\cos(2t).
\label{2-1}
\end{equation}

(a) Write a differential equation for Wronskian of $y_1,y_2,y_3$, which are solutions for homogeneous equation and solve it.

(b) Find fundamental system $\{y_1,y_2,y_3\}$ of solutions for homogeneous equation, and find their Wronskian. Compare with (a).

(c) Find the general solution of (\ref{2-1}).

Boyu Zheng

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Re: TT2B-P2
« Reply #1 on: November 20, 2018, 07:39:28 AM »
Here is my answer
« Last Edit: November 20, 2018, 01:10:21 PM by Boyu Zheng »

Qing Zong

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Re: TT2B-P2
« Reply #2 on: November 20, 2018, 10:23:14 AM »
There are three parts in this question.

Qing Zong

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Re: TT2B-P2
« Reply #3 on: November 20, 2018, 10:25:21 AM »
second part

Qing Zong

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Re: TT2B-P2
« Reply #4 on: November 20, 2018, 10:28:36 AM »
3rd part

Jingze Wang

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Re: TT2B-P2
« Reply #5 on: November 20, 2018, 11:17:19 AM »
a)
The coefficient of $y''$ is -1
then Wronskain is $Ce^{t}$
b)
Use homogeneous equation to find fundamental solutions
$y'''-y''+4y'-4y=0$
Then $r^3-r^2+4r-4=0$
Then $(r^2+4)(r-1)=0$
Then$ r_1=-2i, r_2=2i, r_3=1$
Then the solution is $y=C_1\cos2t+C_2\sin2t+C_3e^t$
So
$W(y_1, y_2, y_3)(t) = \begin{bmatrix} \cos2t&\sin2t&e^{t}\\-2\sin2t&2\cos2t&e^{t}\\-4\cos2t&-4\sin2t&e^{t}\\ \end{bmatrix}=10e^{t}$
This is consistent with what we get in part (a)

c) Use undetermined coefficients method
Assume $y(t) = At\cos{2t}+Bt\sin{2t}$
$y'(t) = A\cos{2t}-2At\sin2t+B\sin2t+2Bt\cos2t$
$y''(t) = -4A\sin{2t}-4At\cos2t+8At\sin2t-8B\sin2t$
$y'''(t) = -12A\cos{2t}+8At\sin2t-12B\sin2t-8Bt\cos2t$

Plug into the equation,

We get $-8A-4B=8, 4A-8B=0$
we get $A=-\frac{4}{5}$
$B=-\frac{2}{5}$

Thus, $Y=C_1\cos2t+C_2\sin2t+C_3e^t-\frac{4}{5}t\cos2t-\frac{2}{5}t\sin2t$


« Last Edit: November 20, 2018, 12:31:31 PM by Jingze Wang »