MAT244--2019F > Term Test 1

Problem 4 (morning)

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Victor Ivrii:
(a) Find the general solution for equation
\begin{equation*}
y'' -6y' +25y =16e^{3x} +102\sin(x) .
\end{equation*}
(b) Find solution, satisfying $y(0)=0$, $y'(0)=0$

Ruojing Chen:
(a)When $$y''-6y'+25y=0$$
$$r^2-6r+25=0$$
$$r=\frac{6\pm\sqrt{6^2-4*25}}{2}=\frac{6\pm\sqrt{-64}}{2}=3\pm4i$$
$$\therefore y_c(x)=c_1e^{3x}Cos(3x)+c_2e^{3x}Sin(3x)$$
When$$y''-6y'+25y=16e^{3x}$$
$$y_p(x)=Ae^{3x}$$
$$y'=3Ae^{3x}$$
$$y''=9Ae^{3x}$$
$$9Ae^{3x}-18Ae^{3x}+25Ae^{3x}=16e^{3x}$$
$$16Ae^{3x}=16e^{3x}$$
$$A=1$$
$$\therefore y_p(x)=e^{3x}$$
When $$y''-6y'+25y=102Sin(x)$$
$$y_p(x)=BCos(x)+CSin(x)$$
$$y'=-BSin(x)+CCos(x)$$
$$y''=-BCos(x)-CSin(x)$$
$$-BCos(x)-CSin(x)+6BSin(x)-6CCos(x)+25BCos(x)+25CSin(x)=102Sin(x)$$
$$Cos(x)(-B-6C+25B)=0$$
$$Sin(x)(-C+6B+25C)=102Sin(x)$$
$$4B-C=0$$
$$4C+B=17$$
$$4B=C, 4(4B)+B=17$$
$$B=1$$
$$C=4$$
$$\therefore y_p(x)=Cos(x)+4Sin(x)$$
$$y(x)=y_c(x)+y_p(x)=c_1e^{3x}Cos(4x)+c_2e^{3x}Sin(4x)+e^{3x}+Cos(x)+4Sin(x)$$
(b) y(0)=0, y(0)=0
$$c_1e^0Cos(0)+c_2e^0Sin(0)+e^0+Cos(0)+4Sin(0)=0$$
$$c_1=-2$$
$$y'=3c_1e^{3x}Cos(4x)-4c_1e^{3x}Sin(4x)+3c_2e^{3x}Sin(4x)+4c_2e^{3x}Cos(4x)+3e^{3x}-Sin(x)+4Cos(x)$$
$$3c_1e^0Cos(0)-4c_1e^0Sin(0)+3c_2e^0Sin(0)+4c_2e^0Cos(0)+3e^0-Sin(0)+4Cos(0)=0$$
$$3c_1+4c_2=-7$$
$$plug in c_1=-2$$
$$3(-2)+4c_2=-7$$
$$c_2=-\frac{1}{4}$$
$$y=-2e^{3x}Cos(4x)-\frac{1}{4}e^3xSin(4x)+e^{3x}+Cos(x)+4Sin(x)$$

OK. V.I.

Yue Sagawa:
Should the homogeneous solution be $𝑦_𝑐(𝑥)=𝑐_1𝑒^{3𝑥}\cos 4𝑥 + 𝑐_2𝑒^{3𝑥}\sin 4𝑥$?

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}

AllanLi:

y’’- 6y'+25y=102sinx+16e^{3x}
first solving the homogenous equation

r^2-6r+25=0
solve for r, we get

r = 3 ± 4i
so we have 𝞴=3 and 𝝻=4. Then we have

y_c=c_1e^{3x}cos4x+c_2e^{3x}sin4x
let

y_p = asinx+bcosx,y_p'=acosx-bsinx,y_p'' = -asinx-bcosx
so we will have

-asinx-bcosx-6(acosx-bsinx)+25(asinx+bcosx)=102sinx
solve for a and b

a=4b,102b = 102
so a =4 , b = 1

y_{p2} = Ce^{3x},y_p2'=3Ce^{3x},y_p2'' = 9Ce^{3x}

9C-18C+25C=16
so we have C=1

y_{p2} = e^{3x}

y= c_1e^{3x}cos4x+c_2e^{3x}sin4x+e^{3x}+4sinx+cosx
next step, we are looking for the derivative of y

y'= c_1(3e^{3x}cos4x-4e^{3x}sin4x)+c_2(3e^{3x}sin4x+4e^{3x}cos4x)+3e^{3x}+4cosx-sinx
since we have y(0)=0, y'(0)=0, so we get 2 functions about the coefficients. By the way, cos0=1 and sin0 =0.

c_1=-2, 3c_1+4c_2+7=0

c_1 =-2 , c_2 = -1/4