Toronto Math Forum

a. Express the general solution of the given system of equations in terms of real-value functions.
$$
x^{\prime}=\left[\begin{array}{rr}{-2} & {1} \\ {1} & {-2}\end{array}\right] x
$$
$$
\begin{aligned} \operatorname{det}(A-\lambda I) &=\left|\begin{array}{cc}{-2-\lambda} & {1} \\ {1} & {-2-\lambda}\end{array}\right| \\ &=(-2-\lambda)(-2-\lambda)-1 \\ &=3+4 \lambda+\lambda^{2} \\ &=(\lambda+3)(\lambda+1) \end{aligned}
$$
$$
\lambda=-3, \lambda=-1
$$

when $\lambda=-3$, $(A-(-3) I) x=0$
$$
\left[\begin{array}{ll|l}{1} & {1} & {0} \\ {1} & {1} & {0}\end{array}\right] \sim\left[\begin{array}{ll|l}{1} & {1} & {0} \\ {0} & {0} & {0}\end{array}\right]
$$
$$
\begin{array}{c}{x_{2}=t, \quad x_{1}=-t} \\ {x=t\left[\begin{array}{c}{-1} \\ {1}\end{array}\right]}\end{array}
$$

when $\lambda=-1$, $(A-(-1) I) x=0$
$$
\left[\begin{array}{rr|r}{-1} & {1} & {0} \\ {1} & {-1} & {0}\end{array}\right] \sim\left[\begin{array}{rr|r}{1} & {-1} & {0} \\ {0} & {0} & {0}\end{array}\right]
$$
$$
\begin{aligned} x_{2} =t &\quad x_{1}=t \\ x=& t\left[\begin{array}{l}{1} \\ {1}\end{array}\right] \end{aligned}
$$

Hence, $y=c_{1} e^{-t}\left[\begin{array}{l}{1} \\ {1}\end{array}\right]+c_{2} e^{-3 t}\left[\begin{array}{c}{-1} \\ {1}\end{array}\right]$

Find the general solution for the equation
$$
y^{\prime \prime}-6 y^{\prime}+8 y=48 \sinh (2 x)
$$
$$
\begin{array}{c}{r^{2}-6 r+8=0} \\ {(r-4)(r-2)=0} \\ {r=4 \quad r=2} \\ {y_c=c_{1} e^{4 x}+c_{2} e^{2 x}}\end{array}
$$
$$
y^{\prime \prime}-6 y^{\prime}+8 y=48 \sinh (2 x)
$$
$$
\begin{array}{l}{\sinh (x)=\frac{e^{x}-e^{-x}}{2}} \\ {\sinh (2 x)=\frac{e^{2 x}-e^{-2 x}}{2}} \\ {48 \sinh (2 x)=24 e^{2 x}-24 e^{-2 x}}\end{array}
$$
$$
y^{\prime \prime}-6 y^{\prime}+8 y=24 e^{2 x}
$$

let $y_{p}(t)=A x e^{2 x}$
$$
y^{\prime}=A e^{2 x}+2 A x e^{2 x} \quad y^{\prime \prime}=4 A e^{2 x}+4 A x e^{2 x}
$$
$$
\begin{aligned} 4 A e^{2 x}+4 A x e^{2 x}-6 A e^{2 x}-12 A x e^{2 x}+8 A x e^{2 x} &=24 e^{2 x} \\ 4 A-6 A &=24 \\ A &=-12 \\ y_{P}(t) &=-12 \times 2^{2 x} \end{aligned}
$$
$$
y^{\prime \prime}-6 y^{\prime}+8 y=-24 e^{-2 x}
$$
Let $y_p(t)=A e^{-2 x} \quad y^{\prime}=-2 A e^{-2 x} \quad y^{\prime \prime}=4 A e^{-2 x}$
$$
\begin{aligned} 44 e^{-2 x}+12 A e^{-2 x}+8 A e^{-2 x} &=-24 e^{-2 x} \\ A &=-1 \\ y_{p} &=-e^{-2 x} \end{aligned}
$$
$$
y=c_{1} e^{4 x}+c_{2} e^{2 x}-12 x e^{2 x}-e^{-2 x}
$$

Verify that the functions $y_1$ and $y_2$ are solutions of the given differential equation. Do they constitute a fundamental set of solutions?
$$
y^{\prime \prime}+4 y=0, y(t)=\cos (2 t), y_{2}(t)=\sin (2 t)
$$
$$
\begin{aligned}
w=\left|\begin{array}{cc}{y_{1}(t)} & {y_{2}(t)} \\ {y_{1}^{\prime}(t)} & {y_{2}^{\prime}(t)}\end{array}\right|&=\left|\begin{array}{cc}{\cos (2 t)} & {\sin (2 t)} \\ {-2 \sin (2 t)} & {2 \cos (2 t)}\end{array}\right|\\
&=2 \cos ^{2}(2 t)+2 \sin ^{2}(2 t) \neq 0
\end{aligned}
$$
$$
\begin{aligned} y_{1}=\cos (2 t) \quad& y_{1}^{\prime}=-2 \sin 2 t \quad y_{1}^{\prime \prime}=-4 \cos 2 t \\ y_{1}^{\prime \prime}+4 y_{1}=&(-4 \cos 2 t)+4 \cos (2 t)=0 \\ y_{2}=\sin (2 t)\quad & y_{2}^{\prime}=2 \cos (2 t) \quad y_{2}^{\prime \prime}=-4 \sin (2 t) \\ y_{2}^{\prime \prime}+4 y_{2} &=-4 \sin (2 t)+4 \sin (2 t)=0 \end{aligned}
$$

Hence, they constitute a fundamental set of solutions.