MAT244-2013F > Quiz 2

Problem 2, night sections

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Victor Ivrii:
Find the general solution of the given differential equation:
\begin{equation*}
y''-y'-2y = -2t + 4t^2.
\end{equation*}

Yangming Cai:

Ka Hou Cheok:
Let the solution $y=y_c+Y$,

The characteristic equation for the homogeneous equation $y''-y'-2y=0$ is
$$r^2-r-2=0$$
Solving the equation we have $r_1=2, r_2=-1$ and hence $$y_c=C_1\exp(2t)+C_2\exp(-t)$$

Let $Y=At^2+Bt+C$, $Y'=2tA+B$, $Y''=2A$.

$Y''-Y'-2Y=(2A)-(2tA+B)-2(At^2+Bt+C)=(-2A)t^2+(-2A-2B)t+(2A-B-2C)=-2t+4t^2$

By comparing the coefficients,
\left\{\begin{aligned} &-2A=4,\\ &-2A-2B=-2,\\ &2A-B-2C=0. \end{aligned}\right.
Then,
\left\{\begin{aligned} &A=-2,\\ &B=3,\\ &C=-7/2. \end{aligned}\right.
So, $Y=-2t^2+3t-\frac{7}{2}$ and hence $$y=y_c+Y=C_1\exp(2t)+C_2\exp(-t)-2t^2+3t-\frac{7}{2}$$

Tianqi Chen:
Question2