### Author Topic: Problem 2, night sections  (Read 4484 times)

#### Victor Ivrii ##### Problem 2, night sections
« 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*}

#### Yangming Cai

• Jr. Member
•  • Posts: 12
• Karma: 7 ##### Re: Problem 2, night sections
« Reply #1 on: October 30, 2013, 08:51:01 PM »

#### Ka Hou Cheok

• Jr. Member
•  • Posts: 5
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• Enjoy the sweet of perfection. ##### Re: Problem 2, night sections
« Reply #2 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 $$y=y_c+Y=C_1\exp(2t)+C_2\exp(-t)-2t^2+3t-\frac{7}{2}$$
« Last Edit: October 31, 2013, 05:35:19 AM by Victor Ivrii »

#### Tianqi Chen

• Newbie
• • Posts: 2
• Karma: 0 ##### Re: Problem 2, night sections
« Reply #3 on: November 01, 2013, 11:24:06 AM »
Question2