Toronto Math Forum
MAT2442013F => MAT244 MathTests => Quiz 2 => Topic started 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*}

answers as follow

Let the solution $y=y_c+Y$,
The characteristic equation for the homogeneous equation $y''y'2y=0$ is
$$r^2r2=0$$
Solving the equation we have $r_1=2, r_2=1$ and hence $$y_c=C_1\exp(2t)+C_2\exp(t)$$
Let $Y=At^2+Bt+C$, $Y'=2tA+B$, $Y''=2A$.
$Y''Y'2Y=(2A)(2tA+B)2(At^2+Bt+C)=(2A)t^2+(2A2B)t+(2AB2C)=2t+4t^2$
By comparing the coefficients,
$$
\left\{\begin{aligned}
&2A=4,\\
&2A2B=2,\\
&2AB2C=0.
\end{aligned}\right.
$$
Then,
$$
\left\{\begin{aligned}
&A=2,\\
&B=3,\\
&C=7/2.
\end{aligned}\right.
$$
So, $Y=2t^2+3t\frac{7}{2}$ and hence $$y=y_c+Y=C_1\exp(2t)+C_2\exp(t)2t^2+3t\frac{7}{2}$$

Question2