MAT244-2018S > Term Test 1

P3-Morning

(1/1)

Victor Ivrii:
(a) Find the general solution for equation
\begin{equation*}
y''(t)-5y'(t)+6y(t)=4e^{t}+e^{2t} .
\end{equation*}

(b) Find solution, satisfying $y(0)=0$, $y'(0)=0$.

Vivian Ngo:
*typed solutions coming soon* (I have class until 9)



*typed solutions coming soon*

Vivian Ngo:
As promised, the typed up solution:

Characteristic Equation:
$r^2-5r+6=0$
$(r-2)(r-3) = 0$
Roots: 3 and 2

Homogeneous Equation: $y_{c}(t)=c_{1}e^{3t}+c_{2}e^{2t}$

Particular solutions:

First particular solution:

$4e^{t}$
$Y(t) = Ae^{t} = Y'(t) = Y''(t)$

$Ae^{t} - 5Ae^{t} + 6Ae^{t} = 4e^{t}$
$A - 5A + 6A = 4$
$A = 2$

Second particular solution:

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


$4Ae^{2t} + 4Ate^{2t} - 5(Ae^{2t} + 2Ate^{2t}) + 6Ate^{2t} = e^{2t}$
$-Ae^{2t} = e^{2t}$
$A = -1$


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

Solving for a solution satisfying $y(0)=0, y'(0) = 0$
$y(0) = c_{1}+c_2 + 2 =0$
$y'(0) = 3c_1 + 2c_2 + 1 = 0$
$c_1 = 2, c_2 = -3$

Thus, $y(t)=2e^{3t} -3e^{2t} + 2e^{t} -te^{2t}$

Victor Ivrii:
Vivian
errors, in part (b)

Meng Wu:
Small Error:
For part(b), $y'(0)$ should be $$y'(0)=3c_1+2c_2+2=0$$
$$\implies\cases{c_1=2\\c_2=-4}$$ and $$y(t)=2e^{3t}-4e^{2t}+2e^t-te^{2t}$$

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