# Toronto Math Forum

## MAT244--2019F => MAT244--Test & Quizzes => Term Test 1 => Topic started by: Victor Ivrii on October 23, 2019, 06:27:03 AM

Title: Problem 4 (afternoon)
Post by: Victor Ivrii on October 23, 2019, 06:27:03 AM
(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 $y(0)=0$, $y'(0)=0$.
Title: Re: Problem 4 (afternoon)
Post by: Hongling Liu on October 23, 2019, 06:45:36 AM
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.
Title: Re: Problem 4 (afternoon)
Post by: Di Qiu on October 23, 2019, 06:51:25 AM
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.
Title: Re: Problem 4 (afternoon)
Post by: Mengyuan Wang on October 23, 2019, 07:53:48 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 (afternoon)
Post by: suyichen on October 23, 2019, 08:45:31 AM
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
$$y(x)=14 e^{-x} \cos (4 x)+e^{-x} \sin (4 x)+2 e^{x}+2 \sin (4 x)-16 \cos (4 x)$$
Title: Re: Problem 4 (afternoon)
Post by: Wang Jingyao on October 23, 2019, 09:52:01 AM
a) $r^2+2r+17=0$

$(r+1)^2=-16$

$r+1=\pm 4 i$

$r_{1}=-1 -\ 4 i$
$r_{2}=-1+\ 4 i$

$y(x)=c_{1} e^{-x} cos{(4x)}+c_{2} e^{-x} sin{(4x)}$

$y^{\prime \prime}+2 y^{\prime}+17 y=40 e^{x}$

$\therefore y_{p}(x)=A e^{x}$

$y_{p}^{\prime}(x)=A e^{x}$

$y_{p}^{\prime \prime}(x)=A e^{x}$

Plug in:

$A e^{x}+2 A e^{x}+17 A e^{x}=40 e^{x}$

$20 A e^{x}=40 e^{x}$

$A=2$

$y_{p}(x)=2 e^{x}$

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

$y_{c}(x)=B \cos 4 x+C \sin 4 x$

$y_{c}^{\prime}(x)=-4 B \sin 4 x+4C \cos 4 x$

$y_{c}^{\prime \prime}(x)=-16 B \cos (4 x)-16 C\sin(4 x)$

Plug in:

$(-16B\cos(4 x)-16C\sin (4 x)+2(-4 B \sin 4 x+4C\cos 4 x)+17(B \cos (4 x)+C\sin (4 x))=130 \sin 4 x$

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

$\left\{\begin{array}{l}{B+8 C=0} \\ {C-8 B=130}\end{array}\right.$

$\left\{\begin{array}{l}{B=-16} \\ {C=2}\end{array}\right.$

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

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

b) $\because~~y(0)=0$ , $y'(0)=0$

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

$y'(x)=-c_{1} e^{-x} cos{(4x)} -{4} c_{1} e^{-x} sin{(4x)}+{4} c_{2} e^{-x} cos{(4x)}-c_2 e^{-x} sin{(4x)}+8 \cos (4 x)+64 \sin (4 x)+2 e^{x}$

Plug in $y(0)=0$:

$0=c_1-16+2$

$\therefore~~ c_{1}=14$

Plug in $y'(0)=0$ and $c_{1}=14$:

$0=-14+4c_2+8+2$

$\therefore~~\left\{\begin{array}{l}{c_{1}=14} \\ {c_{2}=1}\end{array}\right.$

$\therefore~~ y(x)=14 e^{-x} cos{(4x)}+e^{-x}sin{(4x)}-16 \cos (4 x)+2 \sin (4 x)+2 e^{x}$
Title: Re: Problem 4 (afternoon)
Post by: Kunpeng Liu on October 23, 2019, 10:24:04 AM
Hello Mengyuan Wang，your question should be Problem 4(afternoon) rather than Problem 4(morning)
Title: Re: Problem 4 (afternoon)
Post by: Ranran Wang on October 23, 2019, 11:55:59 AM
\textbf{Ans:} let $y=e^{r t}$

$y^{\prime}=r e^{r t}$

$y^{\prime \prime}=r^{2} e^{r t}$

$r^{2} e^{r t}+2 r e^{r t}+17 e^{r t}=0$

$r^{2}+2 r+17=0$

$r=\frac{-2 \pm \sqrt{2^{2}-4 x|x| 7}}{2}=\frac{-2 \pm 8 i}{2}=-1 \pm 4 i=\lambda \pm \mu i$

$\lambda=-1, \quad \mu=4$

$\therefore y=C_{1} e^{\lambda x} \cos (\mu x)+C_{2} e^{\lambda x} \sin (\mu x)=C_{1} e^{-x} \cos (4 x)+C_{2} e^{-x} \sin (4 x)$

Let $y=A e^{x}$

$y^{\prime}=A e^{x}$

$y^{\prime \prime}=A e^{x}$

$A e^{x}+2 A e^{x}+17 A e^{x}=40 e^{x}$

$20 A=40$

$A=2$

$y=2 e^{x}$

Let $y=\operatorname{Asin}(4 x)+B \cos (4 x)$

$y^{\prime}=4 A \cos (4 x)-4 B \sin (4 x)$

$y^{\prime \prime}=-16 A \sin (4 x)-16 B \cos (4 x)$

$-16 A \sin (4 x)-16 B \cos (4 x)+2(4 A \cos (4 x)-4 B \sin (4 x))+17(A \sin (4 x)+B \cos (4 x))=130 \sin (4 x)$

$(-16 A-8 B+17 A) \sin (4 x)+(-16 B+8 A+1713) \cos (4 x)=130 \sin (4 x)$

$(A-8 B) \sin (4 x)+(B+4 A) \cos (4 x)=130 \sin (4 x)$

$\left\{\begin{array}{l}{A-8 B=130} \\ {B+8 A=0}\end{array}\right.$

$\Rightarrow B=-8 A$

$\operatorname{sub} B=-8 A$ into $A-8 B=130$

$A+64 A=130$

$A=2$

$13=-8 x 2=-16$

$\therefore y=2 \sin (4 x)-16 \cos (4 x)$

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

$y^{\prime}=-\cos ^{-x} \cos (4 x)-4 c_{1} e^{-x} \sin (4 x)-c_{2} e^{-x} \sin (4 x)+4\left(2 e^{-x} \cos (4 x)+2 e^{x}+8 \cos (4 x)+64 \sin (4 x)\right.$

$y(0)=0 \Rightarrow x=0, y=0$, then $C_{1} e^{0} \cos (0)+C_{2} e^{0} \sin (0)+2 e^{0}+2 \sin (0)-16 \cos (0)=0$

$C_{1}+2-1=06$

$C_{1}=14$

$y^{\prime}(0)=0$

$-C_{1} e^{0} \cos (0)-4 C_{1} e^{0} \sin (0)-\left(2 e^{0} \sin (0)+4 C_{2} e^{0} \cos (0)+2 e^{0}+8 \cos (0)+64 \sin (0)=0\right.$

$-C_{1}+4 C_{2}+10=0$

$-14+4 C_{2}+10=0$

$C_{2}=1$

$\therefore y=14 e^{-x} \cos (4 x)+e^{-x} \sin (4 x)+2 e^{x}+2 \sin (4 x)-16 \cos (4 x)$
Title: Re: Problem 4 (afternoon)
Post by: Siyan Chen on October 23, 2019, 02:29:18 PM
a) Find the general solution of  $y’’+2y’+17y=40e^x + 130 \sin(4x)$

First, for the complimentary solution, consider the homogeneous equation:
$y’’+2y’+17y=0$

and then, the characteristic equation is: $r^2+2r+17=0$,
$r \ = \ \frac{-2 \pm \sqrt{4-4 \times 17}}{2} \ = \ \frac{-2 \pm 8i}{2} \ = \ -1 \pm 4i$

These roots are a pair of complex conjugates in the form of $\lambda \pm i \mu$,
so the differential equation has a general solution in the form of $y(x)=C_{1} e^{\lambda x} \cos(\mu x)+C_{2} e^{\lambda x} \sin(\mu x)$.

In this case, we have: $y_{c}(x)= e^{-x} (C_{1} \cdot \cos(4x) +C_{2} \cdot \sin(4x))$

Then, for the particular solution, by the method of undetermined coefficients,
suppose $y_{p}=A \cdot e^x$ satisfies the equation: $y’’+2y’+17y=40e^x$

Since, $y_{p}=A \cdot e^x$
$y’_{p}=A \cdot e^x$
$y’’_{p}=A \cdot e^x$

plug into the equation: $A \cdot e^x + 2A \cdot e^x + 17 A \cdot e^x = 20A \cdot e^x = 40e^x$
so, $A=2$
$\therefore y_{p}=2e^x$

For another particular solution, suppose $y_{p}=B \cdot \cos(4x) + C \cdot \sin(4x)$ satisfies the equation: $y’’+2y’+17y=130 \cdot \sin(4x)$

Since, $y_{p}=B \cdot \cos(4x) + C \cdot \sin(4x)$
$y’_{p}=-4B \cdot \sin(4x) + 4C \cdot \cos(4x)$
$y’’_{p}=-16B \cdot \cos(4x) -16 C \cdot \sin(4x)$

plug into the equation:  $-16B \cdot \cos(4x) -16 C \cdot \sin(4x)-8B \cdot \sin(4x) + 8C \cdot \cos(4x) + 17B \cdot \cos(4x) + 17C \cdot \sin(4x) \\ =(B+8C) \cdot \cos(4x) + (C-8B) \cdot \sin(4x) \\ =130 \sin(4x)$

Then, $\begin{cases} C-8B=130 \\ B+8C=0 \end{cases}$

=> $\begin{cases} C=2 \\ B=-16 \end{cases}$

$\therefore y_{p}=-16 \cdot \cos(4x) + 2 \cdot \sin(4x)$

So, the general solution is: $y(x)=e^{-x} (C_{1} \cdot \cos(4x) +C_{2} \cdot \sin(4x))+2e^x-16 \cdot \cos(4x) + 2 \cdot \sin(4x)$

b) when $y(0)=0, y’(0)=0$, and we have
$y’(x)=-e^{-x} (C_{1} \cdot \cos(4x) +C_{2} \cdot \sin(4x))+e^{-x} (-4C_{1} \cdot \sin(4x) +4C_{2} \cdot \cos(4x))+2e^x+64 \cdot \sin(4x) + 8 \cdot \cos(4x)$,

plug $y(0)=0$ into $y(x)$ equation, we get: $0=C_{1}+2-16$ => $C_{1}=14$,

then, plug $y’(0)=0$ into $y’(x)$ equation, we have:
$-C_{1}+4C_{2}+2+8=0 \\ -14+4C_{2}+10=0 \\ C_{2} = 1$

$\therefore y(x)=e^{-x} (-14\cdot \cos(4x) + \cdot \sin(4x))+2e^x-16 \cdot \cos(4x) + 2 \cdot \sin(4x)$
Title: Re: Problem 4 (afternoon)
Post by: Dang Tongbo on October 24, 2019, 01:38:59 PM
(a) r^2 + 2r + 17 = 0
r1 = -1 + 4i
r2 = -1 - 4i
so, y = c1*(e^(-x))*cos(4x) + c2*(e^(-x))*sin(4x))
yp1 = A*(e^x)
y'p1 = A*(e^x)
y''p1 = A*(e^x)
A*(e^x) + 2*A*(e^x) + 17*A*(e^x) = 40*A*(e^x)
A = 2
so yp1 = 2*(e^x)
yp2 = A*sin(4x) + B*cos(4x)
y'p2 = 4*A*cos(4x) - 4*B*sin(4x)
y''p2 = -16*A*sin(4x) -16*B*cos(4x)
-16Asin(4x) -16Bcos(4x) + 8Acos(4x) -8Bsin(4x) + 17Asin(4x) + 17Bcos(4x) = 130sin(4x)
In this way, we can solve that: A = 2 and B = -16
yp2 = 2sin(4x) -16cos(4x)
y = c1*(e^(-x))cos(4x) + c2*(e^(-x))sin(4x) +2*(e^x) + 2sin(4x) - 16cos(4x)
(b) y' = -C1*(e^(-x))cos(4x) -4C1*(e^(-x))sin(4x) -C2*(e^(-x))sin(4x) + 4C2(e^(-x))cos(4x) + 2*(e^x)+ 8cos(4x) + 64sin(4x)
When y(0) = 0, y'(0) = 0,
so, c1 = 14 c2 = 1
y = 14*(e^(-x))cos(4x) +(e^(-x))sin(4x) +2*(e^x) + 2sin(4x)- 16cos(4x)