MAT244--2019F > Term Test 1

Problem 4 (afternoon)

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Wang Jingyao:
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 $c_{1}=14$:

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

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

$\therefore~~ y(x)=14 e^{-x} cos{(4x)}+e^{-x}sin{(4x)}-16 \cos (4 x)+2 \sin (4 x)+2 e^{x}$

Kunpeng Liu:
Hello Mengyuan Wang，your question should be Problem 4(afternoon) rather than Problem 4(morning)

Ranran Wang:
\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 $C_{1} e^{0} \cos (0)+C_{2} e^{0} \sin (0)+2 e^{0}+2 \sin (0)-16 \cos (0)=0$

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

$C_{1}=14$

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

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

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

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

$C_{2}=1$

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

Siyan Chen:
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:
$-C_{1}+4C_{2}+2+8=0 \\ -14+4C_{2}+10=0 \\ C_{2} = 1$

$\therefore y(x)=e^{-x} (-14\cdot \cos(4x) + \cdot \sin(4x))+2e^x-16 \cdot \cos(4x) + 2 \cdot \sin(4x)$

Dang Tongbo:
(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)