Toronto Math Forum

MAT244--2019F => MAT244--Test & Quizzes => Term Test 1 => Topic started by: Victor Ivrii on October 23, 2019, 06:25:04 AM

Title: Problem 4 (morning)
Post by: Victor Ivrii on October 23, 2019, 06:25:04 AM
(a) Find the general solution for equation
\begin{equation*}
y'' -6y' +25y =16e^{3x} +102\sin(x) .
\end{equation*}
(b) Find solution, satisfying $y(0)=0$, $y'(0)=0$
Title: Re: Problem 4 (morning)
Post by: Ruojing Chen on October 23, 2019, 06:36:58 AM
(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.
Title: Re: Problem 4 (morning)
Post by: Yue Sagawa on October 23, 2019, 07:34:01 AM
Should the homogeneous solution be $𝑦_𝑐(𝑥)=𝑐_1𝑒^{3𝑥}\cos 4𝑥 + 𝑐_2𝑒^{3𝑥}\sin 4𝑥$?
Title: Re: Problem 4 (morning)
Post by: Mengyuan Wang on October 23, 2019, 07:54:27 AM

\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}
Title: Re: Problem 4 (morning)
Post by: AllanLi on October 23, 2019, 08:36:55 AM

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

y =-2e^{3x}cos4x-\frac{1}{4}e^{3x}sin4x+e^{3x}+4sinx+cosx
Title: Re: Problem 4 (morning)
Post by: Che Liang on October 23, 2019, 09:29:16 AM
y'' -6y' + 25y = 16e3x + 102sinx
y(0) = 0, y'(0) = 0

(a) The characteristic equation is r2 - 6r +25 = 0,
then (r-3)2 = -16, so r = 3 + 4i or 3- 4i.

so yc(x) = c1e3xcos4x + c2e3xsin4x

Think about y'' -6y' + 25y = 16e3x.
Let y1 = Ae3x,
so y1' = 3Ae3x, y1'' = 9Ae3x.
Substitute them in the equation,
we get 9Ae3x - 18Ae3x +25Ae3x = 16e3x.
So 9A-18A+25A = 16, A = 1.
We can get: y1 = e3x.

Think about y'' -6y' + 25y = 102sinx.
Let y2 = Bcosx + Csinx,
so y2' = -Bsinx +Ccosx,
y2'' = -Bcosx - Csinx.
Substitute them in the equation,
we get -Bcosx - Csinx +6Bsinx - 6Ccosx + 25Bcosx + 25Csinx = 102sinx.
So -B - 6C + 25B =0 and -C + 6B + 25C = 102. Then we know B = 1 and C =4.
We can get: y2 = cosx + 4sinx.

Therefore the general solution is:
y(x) = c1e3xcos4x + c2e3xsin4x + e3x + cosx + 4sinx

(b)
y' = c1(-4e3xsin4x + 3e3xcos4x)
+ c2(4e3xcos4x + 3e3xsin4x)
+ 3e3x - sinx + 4cosx
Substitute  y(0) = 0, y'(0) = 0,
We get 0 = c1 + 0 + 1 + 1 + 0, so c1 = -2.
And   0 = c1(-0 + 3) + c2(4 + 0) + 3 - 0 + 4
so 0 = -6 + 4c2 + 7，c2 = -1/4.

Therefore the solution is:
y(x) = -2e3xcos4x + (-1/4)e3xsin4x + e3x + cosx + 4sinx
Title: Re: Problem 4 (morning)
Post by: Kunpeng Liu on October 23, 2019, 09:35:47 AM
$$Let \, \, \, y=e^{rt}\, \,\, \, {y}'=re^{rt}\, \, \, {y}''=r^{2}e^{rt}\\\\Then\, \, \, \, \, \, r^{2}-6r+25=0\, \, \, \, \, \, \, \\\\\\\\\\r1=\frac{6+\sqrt{36-4\cdot 25\cdot 1}}{2}=3+4i\, \, \, \, \, \, \, r2=\frac{6-\sqrt{36-4\cdot 25\cdot 1}}{2}=3-4i\\\\Yc(x)=C1e^{3x}cos4x+C2e^{3x}sin4x\\\\\\\\\\Let\, \, Yp(x)=Ae^{3x}\, \, \,\, \, \, {y}'=3Ae^{3x}\, \, \,\, \, {y}''=9Ae^{3x}\\\\\\\\\\9Ae^{3x}-18Ae^{3x}+25Ae^{3x}=16e^{3x}\, \, \, \, \, \, 16Ae^{3x}=16e^{3x}\, \, \, \, \, 16A=16\, \,\, \, \, \, A=1\\\\\\\\\\Yp(x)=e^{3x}\\\\\\\\\\let \, \, \, \, Yq(x)=Asinx+Bcosx\, \, ,\, \, {y}'=Acosx-Bsinx\, \, ,\, \, {y}''=-Asinx-Bcosx\\\\\\\\\\-Asinx-Bcosx-6Acosx+6Bsinx+25Asinx+25Bcosx=102sinx\\\\\\\\\\(24A+6B)sinx+(24B-6A)cosx=102sinx\, \, \, \, \, \, 24A+6B=102\, \, \, \, 24B-6A=0\\\\A=4\, \, \, \, B=1\\\\\\\\\\Yq(x)=4sinx+cosx\\\\\\\\\\Y(x)=C1e^{3x}cos4x+C2e^{3x}sin4x+e^{3x}+4sinx+cosx\\\\y(0)=0\\\\\\because y(0)=C1+2=0\, \, \, \, C1=-2\\\\\\\\\\\because\, \, \, \, \, {y}'(0)=0\, \, \, \, {y}'(0)=3C1-0+0+4C2+3+4=0\, \, \, \, C2=-\frac{1}{4}\\\\\\\\\\Y(x)=-2e^{3x}cos4x-\frac{1}{4}e^{3x}sin4x+e^{3x}+4sinx+cosx$$
Title: Re: Problem 4 (morning)
Post by: GuangyuDu on October 23, 2019, 10:01:09 AM
Question 4:
$y''-6y'+25y=16e^{3x}+102\sin (x), y(0)=0, y'(0)=0$

Solution:
$\Gamma^2-6\Gamma+25=0, \Gamma=3\pm 4\pi$
$y_c(x)=c_1e^{3x}\cos (4x)+c_2e^{3x}\sin (4x)$
$y_p(x)=Ae^{3x},y_p'(x)=3Ae^{3x},y_p''(x)=9Ae^{3x}$
$y''-6y'+25y=16Ae^{3x}=16e^{3x}$
$A=1,y_p(x)=e^{3x}$
$y_p(x)=B\cos (x)+C\sin x, y_p'(x)=-B\sin x+C\cos x,$
$y_p''(x)=-B\cos x-C\sin x$
$y''-6y'+25y=102\sin x$
$B=1,C=4,y_p(x)=\cos x+4\sin x$
$y=y_c+y_p=c_1e^{3x}\cos (4x) +c_2e^{3x}\sin (4x)+e^{3x}+\cos x+4\sin x$

Plug in $y(0)=0,y'(0)=0$

we have $c_1=-2,c_2=-\frac14$
$y=-2e^{3x}\cos (4x)-\frac14e^{3x}\sin (4x)+e^{3x}+\cos x+4\sin x$