MAT244-2018S > Term Test 2

TT2--P2

(1/1)

**Victor Ivrii**:

Consider equation

\begin{equation}

y'''-3y'+2y= 18e^{-2t}.

\tag{1}

\end{equation}

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**:

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:

$$r^3-3r+2=0$$

Solving this gives:

$$(r-1)^2(r+2)\rightarrow r_{1}=r_{2}=1, r_{3}=-2$$

Therefore, the homogeneous solution would be:

$$y_{c}(t)=c_{1}e^t+c_{2}te^t+c_{3}e^{-2t}\qquad(2)$$

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:

$$Y(t)=Ae^{-2t}$$

Since $e^{-2t}$ is part of the homogeneous solution, we look for solutions of the form:

$$Y(t)=Ate^{-2t}\qquad(3)$$

Differentiating this:

$$Y'(t)=Ae^{-2t}-2Ate^{-2t} \qquad(4)$$

Differentiating again:

$$Y''(t)=-4Ae^{-2t}+4Ate^{-2t}$$

Differentiating once more:

$$Y'''(t)=12Ae^{-2t}-8Ate^{-2t} \qquad(5)$$

Plugging (3), (4) and (5) into (1):

$$12Ae^{-2t}-8Ate^{-2t}-3Ae^{-2t}+6Ate^{-2t}+2Ate^{-2t}=18e^{-2t}$$

Simplifying gives:

$$9Ae^{-2t}=18e^{-2t}$$

Therefore, $A=2$. Subbing this value of A into (3) and combining it with (2) gives the general solution:

$$y(t)=c_{1}e^t+c_{2}te^t+c_{3}e^{2t}+2te^{-2t}$$

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