# Toronto Math Forum

## MAT244--2019F => MAT244--Test & Quizzes => Quiz-4 => Topic started by: Joy Zhou on October 18, 2019, 09:44:50 PM

Title: TUT0601 Quiz4
Post by: Joy Zhou on October 18, 2019, 09:44:50 PM
Find the general solution of given equation:
$$y^{\prime \prime}+y^{\prime}-6 y=12 e^{3 t}+12 e^{-2 t}$$

Step1. To find the general solution, get the homogeneous equation
$$y^{\prime \prime}+y^{\prime}-6 y=0$$

The characteristic equation is
$$r^{2}+r-6=0$$
$$(r+3)(r-2)=0$$
$$r=2, -3$$

Thus, the solution of homogeneous differential equation $y^{\prime \prime}+y^{\prime}-6 y=0$ is
$$y_{c}(t)=c_{1} e^{-3 t}+c_{2} e^{2 t}$$

Step2. To find the particular solution, use undetermined coefficients method.
Suppose $y_{1}(t)=A e^{3_{t}}$ is a function satisfying the equation
$$y_{1}^{\prime \prime}+y_{1}^{\prime}-6 y_{1}=12 e^{3t}$$

Then,
$$\begin{array}{l}{y_{1}(t)=A e^{3 t}} \\ {y_{1}^{\prime}(t)=3 A e^{3 t}} \\ {y_{1}^{\prime \prime}(t)=9 A e^{3 t}}\end{array}$$
$$9 A e^{3 t}+3 A e^{3 t}-6 A e^{3 t}=12 e^{3 t}$$
$$6 A e^{3 t}=12 e^{3 t}$$
$$A=2$$
So, the solution for $y_{1}(t)$ is
$$y_{1}(t)=2 e^{3 t}$$

Now suppose $y_{2}(t)=B e^{-2 t}$ to satisfies the equation
$$y_{1}^{\prime \prime}+y_{1}^{\prime}-6 y_{1}=12 e^{-2 t}$$

Then,
$$y_{2}(t)=B e^{-2 t}$$
$$y_{2}^{\prime}=-2 B e^{-2 t}$$
$$y_{2}^{\prime \prime}=4 B e^{-2 t}$$
$$4 B e^{-2 t}-2 B e^{-2 t}-6 B e^{-2 t}=12 e^{-2 t}$$
$$-4 B e^{-2 t}=12 e^{-2 t}$$
$$B=-3$$

So, the solution for $y_{2}(t)$ is
$$y_{2}(t)=-3 e^{-2 t}$$

Therefore, the general solution for the non-homogeneous differential equation is
$$y=y_{c}(t)+y_{1}(t)+y_{2}(t)$$
$$y=c_{1} e^{-3 t}+c_{2} e^{2 t}+2 e^{3 t}-3 e^{-2 t}$$