MAT244--2019F > Term Test 1

Problem 4 (afternoon)

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Victor Ivrii:
(a) Find the general solution for equation
\begin{equation*}
y'' +2y' +17 y =40 e^{x} +130\sin(4x) .
\end{equation*}
(b) Find solution, satisfying $y(0)=0$, $y'(0)=0$.

Hongling Liu:
y’’ +2y’ + 17y = 40e^x + 130sin3x
Solution:
a):
r^2 + 2r +17 =0
r1 = -1 + 4i, r2 = -1 -4i
∴y(x) = C1•e^-x•cos4x + C2•e^-x•sin4x
y’’ +2y’ + 17y = 40e^x
Let Yp(x) = A•e^x
Y’ = A•e^x, Y’’ = A•e^x
20•A•e^x = 40•e^x ∴A = 2

Where did you learn this crap? V.I.

Di Qiu:
a)
$$r = \frac{-2\pm\sqrt{4-4\cdot17}}{2} = -1\pm4i$$
$$y_c = c_1e^{-x}cos4x+c_2e^{-x}sin4x$$
$$y_{p1}=Ae^x$$
$$y'_{p1}=Ae^x$$
$$y''_{p1}=Ae^x$$
Therefore, $$Ae^x+2Ae^x+17Ae^x = 40e^x$$
$$A=2$$
$$y_{p1}=2e^x$$
$$y_{p2}=B\sin{4x}+C\cos{4x}$$
$$y'_{p2}=4B\cos{4x}-4C\sin{4x}$$
$$y''_{p2}=-16B\sin{4x}-16C\cos{4x}$$
Therefore, $$-16B\sin{4x}-16C\cos{4x}+ 8B\cos{4x}-8C\sin{4x}+17B\sin{4x}+17C\cos{4x}=130\sin{4x}$$
$$B=2, C=-16$$
$$y_{p2}=2\sin{4x}-16\cos{4x}$$
Then we have:
$$y=c_1e^{-x}cos4x+c_2e^{-x}sin4x+2e^x+2\sin{4x}-16\cos{4x}$$
b)
substitutes $$y(0)=0, y'(0)=0$$ into y and y':
$$c_1+2-16=0$$
$$-c_1+4c_2+2+8=0$$
$$c_1=14, c_2=1$$
$$y=14e^{-x}cos4x+e^{-x}sin4x+2e^x+2\sin{4x}-16\cos{4x}$$

OK. V.I.

Mengyuan Wang:

\begin{array}{c}{y^{\prime \prime}-6 y^{\prime}+25 y=16 e^{3 x}+102 \sin x} \\ {y=e^{7 x}} \\ {y^{\prime}=r{e}^{rx} } \\ {y^{\prime \prime}=r^{2} e^{r x}}\end{array}

\begin{array}{l}{r^{2}-6 r+25=0} \\ {r=3 \pm 4 i} \\ {y=c_{1} e^{3 x} \cos 4 x+c_{2} e^{3 x} \sin 4 x}\end{array}

let

\begin{array}{l}{y=A e^{3x} } \\ {y^{\prime}=3 A e^{3x} } \\ {y^{\prime \prime}=9 A e^{3x}}\end{array}

\begin{array}{rl}{(9 A-18 A+25 A) e^{3x}} & {=16 e^{3x} } \\ {16 A e^{3x}} & {=16 e^{3x} } \\ {A} & {=1} \\ {y} & {=e^{3x} }\end{array}

\begin{aligned} \text { Let } y &=A \sin (x)+B \cos (x) \\ & y^{\prime}=A \cos (x)-B \sin (x) \\ y^{\prime \prime} &=-A \sin (x)-B \cos (x) \end{aligned}

(-A+6 B+25 A) \sin (x)+(-B-6 A+25 B) \cos (x)=102 \sin (x)

\begin{array}{l}{24 A+6 B=102} \\ {24 B-6 A=0}\end{array}

\begin{array}{c}{4 A+B=17} \\ {4 B-A=10}\end{array}

\begin{array}{l}{A=4} \\ {B=1}\end{array}

y=4 \sin (x)+\cos (x)

\begin{array}{l}{y=\operatorname{ces}^{3 x} \cos (x)+c_{2} e^{3 x} \sin (4 x)+e^{3 t}+4 \sin (x)+\cos (x)} \\ {y^{\prime}=3 c_{1} e^{3 x} \cos (4 x)-4 c_{1} e^{3 x} \sin (4 x)+3 \epsilon_{2} e^{3 x} \sin (x)+c_{2} e^{3 x}(x)+3 e^{3 x}+4 \cos x} \\ {y(0)=y^{\prime}(0)=0 \quad C_{1}=-2 \quad C_{2}=-\frac{1}{4}} \\ {y=-2 e^{3 x} \cos (4 x)-\frac{1}{4} e^{3 x} \sin (4 x)+e^{3 x}+\cos (x)+4 \sin (x)}\end{array}

\end{document}

suyichen:
Find the general soluton

(a) $y^{\prime \prime}+2 y^{\prime}+17 y=40 e^{x}+130 \sin (4 x)$

(b) And the solution with $y(0)=0 \quad y^{\prime}(0)=0$

(a)
$$\begin{array}{l}{y^{\prime \prime}+2 y^{\prime}+17 y=0} \\ {\Rightarrow r^{2}+2 r+17=0}\end{array}$$
\begin{aligned} r=\frac{-2 \pm \sqrt{2^{2}-4(1)(17)}}{2(1)} &=\frac{-2 \pm \sqrt{-64}}{2}=\frac{-2 \pm 8 i}{2}=-1 \pm 4 i \\ \lambda=-1 \quad \mu=4 \end{aligned}
$$y_{p}(x)=c_{1} e^{-x} \cos (4 x)+c_{2} e^{-x} \sin (4 x)$$

$$\begin{array}{l}{y^{\prime \prime}+2 y^{\prime}+17 y=40 e^{x}} \\ {y_{1}(x)=A e^{x}, \quad y_{1}^{\prime}(x)=A e^{x}, y_{1}^{\prime \prime}(x)=A e^{x}}\end{array}$$
$$\begin{array}{c}{A e^{x}+2 A e^{x}+17 A e^{x}=40 e^{x}} \\ {20 A e^{x}=40 e^{x}} \\ {A=2}\end{array}$$
$$\therefore y_{1}(x)=2 e^{x}$$
$$\begin{array}{l}{y^{n}+2 y^{\prime}+17 y=130 \sin (4 x)} \\ {y_{2}(x)=B \sin (4 x)+c \cos (4 x)} \\ {y_{2}^{\prime}(x)=4 B \cos (4 x)-4 c \sin (4 x)} \\ {y_{2}^{\prime \prime}(x)=-16 B \sin (4 x)-16 c \cos (4 x)}\end{array}$$

$-16B \sin (4 x)-16 c \cos (4 x)+8B \cos (4 x)-8 c \sin (4 x)+17 B \sin (4 x)+17 c \cos (4 x)=130 \sin (4 x)$

$(-16 B-8 c+17 B) \sin (4 x)+(-16 c+8 B+17 c) \cos (4 x)=130 \sin (4 x)$

$\left\{\begin{array}{l}{B-8 c=130} \\ {8 B+c=0}\end{array}\right.$$\Rightarrow\left\{\begin{array}{l}{B-8 C=130} \\ {64 B+8 C=0}\end{array}\right.$$\begin{array}{rl}{65 B=130} & {B=2} \\ {} & {c=-16}\end{array}$

$$y_{2}(x)=2 \sin (4 x)-16 \cos (4 x)$$

$\therefore$ Tue general solution is
$$y(x)=c_{1} e^{-x} \cos (4 x)+c_{2} e^{-x} \sin (4 x)+2 e^{x}+2 \sin (4 x)-16 \cos (4 x)$$

(b)

\begin{aligned} y^{\prime}(x) =&-c_{1} e^{-x} \cos (4 x)+4 c_{1} e^{-x}(-\sin (4 x))+\left(-c_{2} e^{-x} \sin (4 x)+4 c_{2} e^{-x} \cos (4 x)\right) \\ &+2 e^{x}+8 \cos (4 x)+64 \sin (4 x) \end{aligned}
$$\begin{array}{l}{\text { plug } y(0)=0} \\ {\qquad\Rightarrow c_{1}+2-16=0 \quad C_{1}=14}\end{array}$$
$$\begin{array}{rl}{plug } & {y^{\prime}(0)=0} \\ {\Rightarrow} & {-c_{1}+4 c_{2}+2+8=0}\end{array}$$
$$\begin{array}{c}{-14+4 c_{2}+10=0} \\ {c_{2}=1}\end{array}$$

$\therefore$ The solution with Ivc is
$$y(x)=14 e^{-x} \cos (4 x)+e^{-x} \sin (4 x)+2 e^{x}+2 \sin (4 x)-16 \cos (4 x)$$