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Messages - Ranran Wang

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Quiz-5 / LEC5101 Quiz 5
« on: November 01, 2019, 02:09:40 PM »
5. Verify that the given functions y1 and y2 satisfy the corresponding homogeneous equation; then find a particular solution of the given nonhomogeneous equation.\\

&& (1-t)y''+ty'-y=2(t-1)^{2}e^{-t}, 0<t<1 \\
&& y_{1}(t)=e^{t}, y_{2}(t)=t.

$$(1-t)y''+ty'-y=2(t-1)^{2}e^{-t}, 0<t<1;y_{1}(t)=e^{t}, y_{2}(t)=t$$ 

y_{1}(t)=e^{t} \\
y_{1}'(t)=e^{t} \\

Substitute back into the homogeneous equation:
\ \\
Verified that $y_{1}(t)$ and $y_{2}(t)$ both satisfy the corresponding homogeneous equation.\\
And the complementary solution $y_{c}(t)=c_{1}e^{t}+c_{2}$\\
Now divide both sides of the original equation by $1-t$:\\
\ \\
y_{1}(t) & y_{2}(t) \\
y_{1}'(t) & y_{2}'(t)
Since the particular solution has the form:\\

&&u_{1}(t)=-\int \dfrac{y_{2}(t)g(t)}{W[y_{1},y_{2}](t)}dt\\
&&\quad\quad\ =-\int \dfrac{t\cdot (-2(t-1)e^{-t})}{(1-t)e^{t}}dt\\
&&\quad\quad\ =-2\int te^{-2t}dt\\
&&\quad\quad\ =(t+\dfrac{1}{2})e^{-2t}
&&u_{2}(t)=\int \dfrac{y_{1}(t)g(t)}{W[y_{1},y_{2}](t)}dt\\
&&\quad\quad\ =\int \dfrac{e^{t}\cdot (-2(t-1)e^{-t})}{(1-t)e^{t}}dt\\
&&\quad\quad\ =2\int e^{-t}dt \\
&&\quad\quad\ =-2e^{-t}
$$Y(t)=(t+\dfrac{1}{2})e^{-2t}\cdot e^{t}+(-2e^{-t})\cdot t=(\dfrac{1}{2}-t)e^{-t}$$
Hence, the general solution:\\
Therefore, the particular solution of the given nonhomogeneous equation is\\

Term Test 1 / Re: Problem 4 (afternoon)
« 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 $y=A e^{x}$

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

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

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

$20 A=40$


$y=2 e^{x}$

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 $A-8 B=130$

$A+64 A=130$


$13=-8 x 2=-16$

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

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

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

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




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

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

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


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

Term Test 1 / Re: Problem 3 (afternoon)
« on: October 23, 2019, 11:55:19 AM »
\textbf{Q3. $y^{\prime \prime}-5 y^{\prime}+6 y=52 \cos (2 x)$,$y(0)=0$,$y^{\prime}(0)=0$}

\textbf{Ans:} let $y=e^{r x}$

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

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

$r^{2} e^{r x}-5 r e^{r x}+6 e^{r x}=0$

$\left(r^{2}-5 r+6\right)=0$




$y=a e^{r x}+c_{2} e^{r_{2} x}=u e^{2 x}+c_{2} e^{3 x}$

let $y=A \cos (2 x)+B \sin (2 x)$

$y^{\prime}=-2 A \sin (2 x)+2 B \cos (2 x)$

$y^{\prime \prime}=-4 A \cos (2 x)-4 B \sin (2 x)$

$y^{\prime \prime}-5 y^{\prime}+6 y=52 \cos (2 x)$

$-4 A \cos (2 x)-4 B \sin (2 x)-5(-2 A \sin (2 x)+2 B \cos (2 x))+6(A \cos (2 x)+B \sin (2 x))=52(\sin (2 x)$

$(-4 A-10 B+6 A) \cos (2 x)+(-4 B+10 A+6 B) \sin (2 x)=52 \cos (2 x)$

$(-10 B+2 A) \cos (2 x)+(10 A+2 B) \sin (2 x)=52 \cos (2 x)$

$\left\{\begin{array}{c}{-10 B+2 A=52} \\ {10 A+2 B=0}\end{array}\right.$ $\Rightarrow 5 A+B=0$

$\Rightarrow \quad B=-5 A$  Sub $B=-5 A$ into $-10 B+2 A=52$

$50 A+2 A=52$



$\therefore y=\cos (2 x)-5 \sin (2 x)$

$\therefore y=c_{1} e^{2 x}+c_{2} e^{3 x}+\cos (2 x)-5 \sin (2 x)$

then $y^{\prime}=2 C_{1} e^{2 x}+3 C_{2} e^{3 x}-2 \sin (2 x)-10 \cos (2 x)$

$y(0)=0$  sub $x=0, \quad y=0$

$0=C_{1} e^{0}+\left(2 e^{0}+\cos (0)-5 \sin (0)\right)=C_{1}+C_{2}+1=0$


$y^{\prime}(0)=0$  sub $x=0, y^{\prime}=0$
$0=2C_{1}  e^{0}+3\left(2 e^{0}-2 \sin (0)-10 \cos (0)=2 C_{1}+3 C_{2}-10=0\right.$

$2C_{1}+3 C_{2}=10$

Sub $c_{2}=-1-C_{1}$ info $2C_{1}+3 C_{2}=10$

$2C_{1}+\left(-3-3 C_{1}\right)=10$

$2 C_{1}-3-3 C_{1}=10$



$\therefore y=-13 e^{2 x}+12 e^{3 x}+\cos (2 x)-5 \sin (2 x)$

Term Test 1 / Re: Problem 2 (afternoon)
« on: October 23, 2019, 11:53:55 AM »
\textbf{Q2. }$(2 x+1) x y^{\prime \prime}+(2 x+2) y^{\prime}-2 y=0$.  Fnd $w\left(y_{1}, y_{2}\right)$. Check $y_{1}(x)=x+1$ is $a$ solution and fnd another linearly independent solution. Write general solution, and find solutions such fhat $y(-1)=1, \quad y^{\prime}(-1)=0$.

 \textbf{Ans:} $y^{\prime \prime}+\frac{2 x+2}{(2 x+1) x} y^{\prime}-\frac{2}{(2 x+1) x} y=0$  Fmd $P(x)=\frac{2 x+2}{(x+1) x}=\frac{A}{2 x+1}+\frac{B}{x}=\frac{A x+2 B x+B}{(2 x+1) x}$

$W=C e^{\left.-\int p( x\right) d x}=Ce^{-\int\left(\frac{-2}{2 x+1}+\frac{2}{x}\right) d x}=C e^{\int\left(\frac{2}{2 x+1}-\frac{2}{x}\right) d x}={Ce}^{\int \frac{2}{2 x+1} d x-\int \frac{2}{x} d x}=\operatorname{Ce}^{\ln| 2 x+1|-2 \ln | x |}=c \frac{e^{\ln |2 x+1|}}{e^{\sin |x|}}$

$W=\frac{2 x+1}{x^{2}}=\left|\begin{array}{ll}{y_{1}(x)} & {y_{2}(x)} \\ {y_{1}^{\prime}(x)} & {y_{2}^{\prime}(x)}\end{array}\right|=\left|\begin{array}{cc}{x+1} & {y_{2}(x)} \\ {1} & {y^{\prime}(x)}\end{array}\right|=(x+1) y_{2}^{\prime}(x)-y_{2}(x)$



$y_{1}^{\prime \prime}(x)=0$

$(2 x+1) x \cdot 0+(2 x+2) \cdot 1-2(x+1)=2 x+2-2 x-2=0$

$y_{2}^{\prime}(x)-\frac{1}{x+1} y_{2}(x)=\frac{2 x+1}{x^{2}(x+1)}$


$\mu=e^{\int p(x) d x}=e^{-\int \frac{1}{x+1} d x}=e^{-\ln |x+1|}=\frac{1}{x+1}$

$\frac{1}{x+1} y_{2}^{\prime}(x)-\frac{1}{(x+1)^{2}} y_{2}(x)=\frac{2 x+1}{x^{2}(x+1)^{2}}$

$\left[\frac{1}{x+1} y_{2}(x)\right]^{\prime}=\int \frac{2 x+1}{\left(x^{2}+x\right)^{2}} d x$

$\frac{1}{x+1} y_{2}(x)=\frac{-1}{x^{2}+x}$


$y=C_{1} y_{1}+C_{2} y_{2}=C_{1}(x+1)+C_{2} \frac{-1}{x}=C_{1}(x+1)+C_{2} \frac{-1}{x}$

$y(-1)=1, \quad y=1, \quad x=-1$

$C_{1}(-1+1)+C_{2} \frac{-1}{-1}=1$










Term Test 1 / Re: Problem 1 (afternoon)
« on: October 23, 2019, 11:52:18 AM »

$M=-y^{2} \sin (x y)$

$N=-x y \sin (x y)+2 \cos (x y)+3 y$

$M_{y}=-2 y \sin (x y)-y^{2} \cos (x y) \cdot x=-2 y \sin (x y)-x y^{2} \cos (x y)$

$N_{x}=-y \sin (x y)-x y^{2} \cos (x y)+2(-\sin (x y)) y+0=-y \sin (x y)-x y^{2}(\cos (x y)-2 y \sin (x y)$

$R_1=\frac{M_{y}-N_{x}}{M}=\frac{-2 y \sin (x y)-x y^{2} \cos (x y)+y \sin (x y)+x y^{2} \cos (x y)+2 y \sin (x y)}{-y^{2} \sin (x y)}=\frac{y \sin (x y)}{-y^{2} \sin (x y)}=-\frac{1}{y}$

$\mu=e^{-\int_{R_1} d y}=e^{\int \frac{1}{y} d y}=e^{\ln |y|}=y$

both muliply by $\mu$ $-y^{3} \sin (x y)+\left(-x y^{2} \sin (x y)+2 y \cos (x y)+3 y^{2}\right) y^{\prime}=0$

$\varphi(x, y)$

$\varphi_{x}=M=-y^{3} \sin (x y)$

$\varphi_{y}=N=-x y^{2} \sin (x y)+2 y \cos (x y)+3 y^{2}$

$\varphi=\int M d x=\int-y^{3} \sin (x y) d x=y^{2} \cos (x y)+h(y)$

$\varphi_{y}=2 y \cos (x y)+y^{2} x \sin (x y)+h^{\prime}(y)=N$

$h^{\prime}(y)=3 y^{2}$

$\int h^{\prime}(y) d y=\int 3 y^{2} d y=y^{3}$

$\varphi=y^{2} \cos (x y)+y^{3}=C$

put $y\left(\frac{\pi}{3}\right)=1$


$y=1$  into $ \varnothing $

$1^{2} \cos \left(\frac{\pi}{3} \times 1\right)+1^{3}=C=\frac{1}{2}+1=\frac{3}{2}$

$\therefore y^{2} \cos (x y)+y^{3}=\frac{3}{2}$

Quiz-4 / TUT0501 Quiz4
« on: October 18, 2019, 09:44:52 PM »
This is my solution for the question in quiz.

Quiz-3 / TUT 0501 Quiz 3
« on: October 11, 2019, 03:33:38 PM »
Here is my question and solution for the question I met in Quiz 3.

Quiz-2 / TUT0501 Quiz2
« on: October 04, 2019, 03:22:06 PM »
This is the question I do in my quiz 2 and my solution for the question.

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