Author Topic: TT2--P2  (Read 5484 times)

Victor Ivrii

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« on: March 21, 2018, 02:56:30 PM »
Consider equation
y'''-3y'+2y= 18e^{-2t}.
a.  Write 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 (1).

Jared Jubas-Malz

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Re: TT2--P2
« Reply #1 on: March 21, 2018, 09:08:55 PM »
Part (a)$\\$
The equation for the Wronskian would be:
$$W = c\times exp[-\int p_{1}(t)dt] $$
Since there is no $y''$ term, $p_{1}(t)$ would be $0$:
$$W = c\times exp[-\int 0 dt]=c\times exp(0)=c$$
Therefore, the Wronskian would be a constant.

Part (b)$\\$
Consider the homogeneous equation:
$$y''' - 3y' + 2y = 0$$
The characteristic equation would be:
Solving this gives:
$$(r-1)^2(r+2)\rightarrow r_{1}=r_{2}=1, r_{3}=-2$$
Therefore, the homogeneous solution would be:
Computing the Wronskian:
$$W=\begin{array}{|c c c|}e^t &te^t &e^{-2t}\\e^t&(t+1)e^t&-2e^{-2t}\\e^t&(t+2)e^t&4e^{-2t}\end{array}=4(t+1)+2(t+2)-t(4+2)+(t+2)-(t+1)=9$$
Therefore, the Wronskian is a constant just as expected based on part (a).

Part (c)$\\$
The particular solution should be of the form:
Since $e^{-2t}$ is part of the homogeneous solution, we look for solutions of the form:
Differentiating this:
$$Y'(t)=Ae^{-2t}-2Ate^{-2t} \qquad(4)$$
Differentiating again:
Differentiating once more:
$$Y'''(t)=12Ae^{-2t}-8Ate^{-2t} \qquad(5)$$
Plugging (3), (4) and (5) into (1):
Simplifying gives:
Therefore, $A=2$. Subbing this value of A into (3) and combining it with (2) gives the general solution:
« Last Edit: March 21, 2018, 11:07:43 PM by Jared Jubas-Malz »