MAT244--2019F > Term Test 1

Problem 3 (morning)

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Victor Ivrii:
(a) Find the general solution for equation
\begin{equation*}
y'' -6y'+8 y= 48\sinh (2x).
\end{equation*}
(b) Find solution, satisfying $y(0)=0$, $y'(0)=0$.

Hint: $\sinh(x)=\dfrac{e^x-e^{-x}}{2}$.

Ruojing Chen:
As$$Sinh(2x)=\frac{e^{x}-e^{-x}}{2}$$
$$48Sinh(2x)=48(\frac{e^{2x}-e^{-2x}}{2})=24e^{2x}-24e^{-2x}$$
when $$y''-6y'+8y=0$$ $$r^2-6r+8=0$$
$$(r-4)(r-2)=0$$
$$r_1=4,r_2=2$$
$$\therefore y_c(x)=c_1e^{4x}+c_2e^{2x}$$
when $$y''-6y'+8=24e^{2x}$$
$$y_p(x)=Ae^{2x}x$$
$$y'=2Ae^{2x}x+Ae^{2x}$$
$$y''=4Ae^{2x}x+2Ae^{2x}+2Ae^{2x}=4Ae^{2x}x+4Ae^{2x}$$
$$4Ae^{2x}x+4Ae^{2x}-12Ae^{2x}x-6Ae^{2x}+8Ae^{2x}x=24e^{2x}$$
$$-2Ae^{2x}=24e^(2x)$$
$$A=-12$$
$$\therefore y_p(x)=-12e^{2x}x$$
when$$y''-6y'+8=-24e^{-2x}$$
$$y_p(x)=Be^{-2x}$$
$$y'=-2Be^{-2x}$$
$$y''=4Be^{-2x}$$
$$4Be^{-2x}+12Be^{-2x}+8Be^{-2x}=-24e^{-2x}$$
$$24Be^{-2x}=-24e^{-2x}$$
$$B=-1$$
$$\therefore y_p(x)=-e^{-2x}$$

$$y=y_c(x)+y_p(x)=c_1e^{4x}+c_2e^{2x}-12e^{2x}x-e^{-2x}$$

(b)when y(0)=0 $$c_1e^0+c_2e^0-12e^o*0-e^0=0$$
$$c_1+c_2=1$$
when y'(0)=0
$$y'=4c_1e^{4x}+2c_2e^{2x}-24e^{2x}x-12e^{2x}+2e^{-2x}$$
$$4c_1e^0+2c_2e^0+24e^0*0-12e^0+2e^0=0$$
$$4c_1+2c_2-12+2=0$$
$$2c_1+c_2=5$$
$$\therefore c_1=4,c_2=-3$$

$$y=4e^{4x}-3e^{2x}+12e^{2x}x-e^{-2x}$$ Some errors. V.I.

Mengyuan Wang:

\begin{array}{l}{y^{\prime \prime}-6 y^{\prime}+8 y=48 \sinh (2 x)} \\ {y^{\prime \prime}-6 y^{\prime}+8 y=24 e^{2 x}-24 e^{-x}}\end{array}

\text { Let } y=e^{r x}

\begin{array}{l}{y^{\prime}=\sec ^{r x}} \\ {y^{\prime \prime}=r^{2} e^{r x}}\end{array}

\begin{array}{c}{y^{2}-6 x+8=0} \\ {r_{1}=2 \quad r_{2}=4} \\ {y=c_{1} e^{2x} +c_{2} e^{4x} }\end{array}

\text { let } y=A x e^{2 x}

y^{\prime}=A e^{2 x}+2 A x e^{2 x}

y^{\prime \prime}=4 A xx^{2 x}+4 A e^{2 x}

\begin{array}{c}{(8 A-12 A+4A) t \cdot e^{2 x}+(-6 A+4 A) e^{2 x}=24 e^{2 x}} \\ {-2 Ae^{2 x}=24 e^{2 x}} \\ {A=-12} \\ {y=-12 x e^{2 x}}
\end{array}

Let

\begin{array}{l}{y=A e^{-2 x}} \\ {y^{\prime}=-2 A e^{-2 x}} \\ {y^{\prime \prime}=4 A e^{-2 x}}\end{array}

\begin{aligned}(8 A+12 A+4 A) e^{-2 x} &=-24 e^{-2 x} \\ A &=-1 \\ y &=e^{-2 x} \end{aligned}

so

\begin{array}{l}{y=a e^{2}+c_{2} e^{4 x}-12 e^{2 x}-e^{-2 x}} \\ {y^{\prime}=2 c_{1} e^{2 t}+4 c_{2} e^{4 t}-24 r e^{2 x}+2 e^{-2 x}-12 e^{2 x}} \\ {y(0)=y^{\prime}(0)=0} \\ {y=-3 e^{2 x}+4 e^{4 x}-12 x e^{2 x}-e^{-2 x}}\end{array}

OK. V.I.

Yiyang Huang:
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}
$$y=c_{1} e^{4 x}+c_{2} e^{2 x}-12 x e^{2 x}-e^{-2 x}$$

AllanLi:

y''-6y'+8y=48sinh(2x)
applying the hint

y''- 6y' + 8y= 24e^{2x}-24e^{-2x}
solving the homogeneous equation

r^2-6r+8=0 ,  r_1 = 2, r_2 = 4

Yc = C_1e^2 + C_2e^4
let Yp = Axe^{2x}

Yp' = A(e^{2x} + 2xe^{2x}), Yp'' = A(2e^2x + 2(e^{2x} +2xe^{2x}))
fill in the original equation

4Ae^{2x}+4Axe^{2x} - 6A(e^{2x}+2xe^{2x})+8Axe^{2x}=24e^{2x}
solve for A

A = -12
let

Ys = Be^{-2x}

Ys' = -2Be^{-2x} , Ys'' = 4Be^{-2x}

4Be^{-2x} -6(-2Be^{-2x}) + 8Be^{-2x} = -24e^{-2x}
then we have

24B = -24, B = -1
so we have general solution

y = c1e^{2x} + c2e^{4x} -12xe^{2x}-e^{-2x}
Part(b)

y(0)=0, y'(0)=0
we get

C_1+C_2-1 =0 , 2C_1 + 4C_2 -12+2 = 0
solve for C1 and C2

C_1 = -3 , C_2 = 4