# Toronto Math Forum

## MAT244-2013F => MAT244 Math--Tests => Quiz 2 => Topic started by: Victor Ivrii on October 30, 2013, 08:11:06 PM

Title: Problem 2, night sections
Post by: Victor Ivrii on October 30, 2013, 08:11:06 PM
Find the general solution of the given differential equation:
\begin{equation*}
y''-y'-2y = -2t + 4t^2.
\end{equation*}
Title: Re: Problem 2, night sections
Post by: Yangming Cai on October 30, 2013, 08:51:01 PM
Title: Re: Problem 2, night sections
Post by: Ka Hou Cheok on October 30, 2013, 08:54:48 PM
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 $$y=y_c+Y=C_1\exp(2t)+C_2\exp(-t)-2t^2+3t-\frac{7}{2}$$
Title: Re: Problem 2, night sections
Post by: Tianqi Chen on November 01, 2013, 11:24:06 AM
Question2