# Toronto Math Forum

## MAT244-2018S => MAT244--Tests => Term Test 1 => Topic started by: Victor Ivrii on February 15, 2018, 05:08:16 PM

Title: P4-Day
Post by: Victor Ivrii on February 15, 2018, 05:08:16 PM
Find the general solution for equation
\begin{equation*}
y''(t)-4y'(t)+5y(t)=2 e^{t}+ 8\cos(t).
\end{equation*}
Title: Re: P4-Day
Post by: Meng Wu on February 15, 2018, 10:35:21 PM
$(a)$ $\\$

First find the complementary solution for the homogeneous equation:
$$y’’-4y’+5y=0$$
Characteristic equation: $$r^2-4r+5=0 \implies \cases{r_1=2+i\\r_2=2-i}$$
Thus $$y_c(t)=c_1e^{2t}cos(t)+c_2e^{2t}sin(t)$$
Now we need to find the particular solution:
$$y_p(t)=Y_1(t)+Y_2(t)$$
We assume $Y_1(t)=Ae^t$ and there are no duplicates of $y_c(t)$,
$\\$thus $Y_1’(t)=Ae^t$ and $Y_1’’(t)=Ae^t$. $\\$
Substitue theses values back to $y’’-4y’+5y=2e^t$:
$$Ae^t-4Ae^t+5Ae^t=2e^t \implies A=1$$
We assume $Y_2(t)=Bcos(t)+Csin(t)$ and there are no duplicates of $y_c(t)$, $\\$thus $Y_2’(t)=-Bsin(t)+Ccos(t)$ and $Y_2’’(t)=-Bcost(t)-Csin(t)$. $\\$
Substitute these values back to $y’’-4y’+5y=8cos(t)$:
$$-Bcost(t)-Csin(t)+4Bsin(t)-4Ccos(t)+5Bcos(t)+5Csin(t)=8cos(t) \implies \cases{B=1\\C=-1}$$
Thus, $$y_p(t)=Y_1(t)+Y_2(t)=e^t+cos(t)-sin(t)$$
Therefore, the general solution is \begin{align}y(t)&=y_c(t)+y_p(t)\\&=c_1e^{2t}cos(t)+c_2e^{2t}sin(t)+e^t+cos(t)-sin(t)\end{align}

$(b)$ $\\$
$$y’(t)=2c_1e^{2t}cos(t)-c_1e^{2t}sin(t)+2c_2e^{2t}sin(t)+c_2e^{2t}cos(t)+e^t-sin(t)-cos(t)$$
Set $t=0$ and $y=0$; $t=0$ and $y’(t)=0$:
$$\cases{c_1+0+1+1-0=0\\2c_1-0+0+c_2+1-0-1=0} \implies \cases{c_1=-2\\c_2=4}$$
Therefore, the solution for the IVP is $$y(t)=-2e^{2t}cos(t)+4e^{2t}sin(t)+e^t+cos(t)-sin(t)$$