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### Messages - Linqian Shen

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1
##### Quiz-5 / LEC0101 quiz5
« on: November 01, 2019, 03:02:29 PM »
Find the general solution of the given differential equation.
$$y^{\prime \prime}+y=\tan (t)$$

$$\begin{array}{c}{r^{2}+1=0} \\ {r=\pm i} \\ {y=c_{1} \cos t+c_{2} \sin t}\end{array}$$
\begin{aligned} w&=\left|\begin{array}{cc}{\cos t} & {\sin t} \\ {-\sin t} & {\cos t}\end{array}\right|\\ &=\cos ^{2} t+\sin ^{2} t=1\\ w_{1}&=\left|\begin{array}{ll}{0} & {\sin t} \\ {1} & {\cos t}\end{array}\right|=-\sin t\\ w_{2}&=\left|\begin{array}{cc}{\cos t} & {0} \\ {-\sin t} & {1}\end{array}\right|=\cos t \end{aligned}
\begin{aligned} y p(t)&=\cos t \int \frac{-s i n t-\tan s}{1} d s+s \sin t \int \frac{\cos s-t a n s}{1} d s\\ &=\cos t \int-\sin s \frac{\sin s}{\cos s} d s+\sin t \int \cos s \frac{\sin s}{\cos s} d s\\ &=-\cos t \int \frac{1-\cos ^{2} s}{\cos s} d s+\sin t \int \sin s d s\\ &=-\cos t \int \sec -\cos s d s+\sin t(-\cos s)\\ &=-\cos \ln (\sec t+\tan t)+\cos t \sin t-\sin t \cos t \end{aligned}
$$y(t)=c_{1} \cos t+c_{2} \sin t-\cos t \ln (\sec t+\tan t)$$

2
##### Term Test 1 / Re: Problem 3 (afternoon)
« on: October 23, 2019, 08:02:11 AM »
Question:

find a general solution of $y^{\prime \prime}-5 y^{\prime}+6 y=52 \cos (2 x)$
and satisfy $y(0)=0, y^{\prime}(0)=0$
$$\begin{array}{c}{y^{\prime \prime}-5 y^{\prime}+6 y=0} \\{(r-2)(r-3)=0}\end{array}$$
$$\begin{array}{l}{r_{1}=2, r_{2}=3} \\ {y_{c}(t)=c_{1} e^{2 t}+c_{2} e^{3 t}}\end{array}$$
$$\begin{array}{l}\\ {y_{p}(t)=A \cos (2 x)+B \sin (2 x)} \\ {y^{\prime} p(t)=-2 A \sin (2 x)+2 B \cos (2 x)} \\ {y^{\prime \prime} p(t)=-4 A \cos (2 x)-4B \sin (2 x)}\end{array}$$
$$\begin{array}{rl}{-4 A \cos (2 x)-4 B \sin (2 x)+10 A \sin (2 t)-10} & {B \cos (2 x)+6 A \cos (2 x)+6 B \sin (2 x)} \\ {} & {=52 \cos (2 x)}\end{array}$$
$$(-4 A-10 B+6 A) \cos (2 x)+(-4 B+10 A+6 B) \sin (2 x)=52 \cos (2 x)$$
$$\left\{\begin{array}{l}{2 A-10 B=52} \\ {2 B+10 A=0}\end{array} \quad \left\{\begin{array}{l}{A=1} \\ {B=-5}\end{array}\right.\right.$$
$$\begin{array}{l}{y_p(x)=\cos (2 x)-5 \sin (2 x)} \\ {y(x)=y_c(x)+y_p(x)=c_{1} e^{2 t}+c_{2} e^{3 t}+\cos (2 x)-5 \sin (2 x)}\\{ y^{\prime}(x)=2 c_1 e^{2 t}+3 c_{2} e^{3 t}-2 \sin (2 x)-10 \cos (2 x)} \\ { y(0)=0, y^{\prime}(0)=0}\\{\therefore\left\{\begin{array}{l}{c_{1}+c_{2}=-1} \\ {2 c_{1}+3 c_{2}=10}\end{array}\therefore \left\{\begin{array}{l}{c_{1}=-13} \\ {c_{2}=12}\end{array}\right.\right.}\\{\therefore y(x)=-13 e^{2 t}+12 e^{3 t}+\cos (2 x)-5 \sin (2 x)} \end{array}$$

3
##### Term Test 1 / Problem3 (afternoon)
« on: October 23, 2019, 08:00:14 AM »
Question:

find a general solution of $y^{\prime \prime}-5 y^{\prime}+6 y=52 \cos (2 x)$
and satisfy $y(0)=0, y^{\prime}(0)=0$
$$\begin{array}{c}{y^{\prime \prime}-5 y^{\prime}+6 y=0} \\{(r-2)(r-3)=0}\end{array}$$
$$\begin{array}{l}{r_{1}=2, r_{2}=3} \\ {y_{c}(t)=c_{1} e^{2 t}+c_{2} e^{3 t}}\end{array}$$
$$\begin{array}{l}\\ {y_{p}(t)=A \cos (2 x)+B \sin (2 x)} \\ {y^{\prime} p(t)=-2 A \sin (2 x)+2 B \cos (2 x)} \\ {y^{\prime \prime} p(t)=-4 A \cos (2 x)-4B \sin (2 x)}\end{array}$$
$$\begin{array}{rl}{-4 A \cos (2 x)-4 B \sin (2 x)+10 A \sin (2 t)-10} & {B \cos (2 x)+6 A \cos (2 x)+6 B \sin (2 x)} \\ {} & {=52 \cos (2 x)}\end{array}$$
$$(-4 A-10 B+6 A) \cos (2 x)+(-4 B+10 A+6 B) \sin (2 x)=52 \cos (2 x)$$
$$\left\{\begin{array}{l}{2 A-10 B=52} \\ {2 B+10 A=0}\end{array} \quad \left\{\begin{array}{l}{A=1} \\ {B=-5}\end{array}\right.\right.$$
$$\begin{array}{l}{y_p(x)=\cos (2 x)-5 \sin (2 x)} \\ {y(x)=y_c(x)+y_p(x)=c_{1} e^{2 t}+c_{2} e^{3 t}+\cos (2 x)-5 \sin (2 x)}\\{ y^{\prime}(x)=2 c_1 e^{2 t}+3 c_{2} e^{3 t}-2 \sin (2 x)-10 \cos (2 x)} \\ { y(0)=0, y^{\prime}(0)=0}\\{\therefore\left\{\begin{array}{l}{c_{1}+c_{2}=-1} \\ {2 c_{1}+3 c_{2}=10}\end{array}\therefore \left\{\begin{array}{l}{c_{1}=-13} \\ {c_{2}=12}\end{array}\right.\right.}\\{\therefore y(x)=-13 e^{2 t}+12 e^{3 t}+\cos (2 x)-5 \sin (2 x)} \end{array}$$

4
##### Quiz-4 / TUT0402 quiz4
« on: October 18, 2019, 10:18:57 PM »
Find the general solution of the given differential equation
$$14 \cdot y^{\prime \prime}-y^{\prime}-2 y=\cosh (2 t)=\frac{1}{2} e^{2 t}+\frac{1}{2} e^{-2 t}$$

$$\begin{array}{c}{r^{2}-r-2=0} \\ {(r+1)(r-2)=0} \\ {r_{1}=-1 \quad r_{2}=2}\end{array}$$
$$y=c_{1} e^{-t}+c_{2} e^{2 t}$$

$$y_{1}(t)=A e^{2 t}$$
$$y_{1}(t)=A t e^{2 t}$$
$$\begin{array}{l}{y^{\prime}(t)=2 A t e^{2 t}+A e^{2 t}} \\ {y^{\prime \prime}(t)=4 A t e^{2 t}+2 A e^{2 t}+2 A e^{2 t}} \\ {4 A t e^{2 t}+2 A e^{2 t}+2 A e^{2 t}-2 A t e^{2 t}-A e^{2 t}-2 A t e^{2 t}=\frac{1}{2} e^{2 t}} \\ {e^{2 t}(4 A t+2 A+2 A-2 A t-A-2 A+)=\frac{1}{2} e^{2 t}}\end{array}$$
$$\begin{array}{l}{3 A=\frac{1}{2} \quad A=\frac{1}{6}} \\ {y_{1}(t)=\frac{1}{6}+e^{2 t}}\end{array}$$

$$\begin{array}{l}{y_{2}(t)=B e^{-2 t} \quad y_{2}^{\prime}(t)=-2 B e^{-2 t}} \\ {y_{2}^{\prime \prime}(t)=4 B e^{-2 t}}\end{array}$$
$$\begin{array}{c}{4 B e^{-2 t}+2 B e^{-2 t}-2 B e^{2 t}=\frac{1}{2} e^{-2 t}} \\ {4 B e^{-2 t}=\frac{1}{2} e^{-2 t}} \\ {4 B=\frac{1}{2}}\end{array}$$
$$B=\frac{1}{8} \quad y_{2}(t)=\frac{1}{8} e^{-2 t}$$
$$y(t)=c_{1} e^{-t}+c_{2} e^{2 t}+\frac{1}{6} t e^{2 t}+\frac{1}{8} e^{-2 t}$$

5
##### Quiz-3 / TUT0402 quiz3
« on: October 11, 2019, 02:00:01 PM »
Find the Wrouskian of the given parts of function
$$\cos ^{2}(x) \quad 1+\cos (2 x)$$
\begin{aligned} w&=\operatorname{det}\left|\begin{array}{cc}{\cos ^{2}(x)} & {1+\cos (2 x)} \\ {-2 \cos (x) \sin (x)} & {-2 \sin (2 x)}\end{array}\right|\\ &=-2 \sin (2 x) \cos ^{2}(x)+[2 \cos (x) \sin (x)][1+\cos (2 x)]\\ &=-2 \sin (2 x) \cos ^{2}(x)+2 \cos (x) \sin (x)+2 \cos (x) \sin (x) \cos (2 x)\\ &=-2 \sin (2 x) \cos ^{2}(x)+\sin (2 x)+\sin (2 x) \cos (2 x)\\ &=\sin (2 x)\left(-2 \cos ^{2}(x)+1+\cos (2 x)\right)\\ &=\sin (2 x)\left(1-2 \cos ^{2}(x)+2 \cos ^{2}(x)-1\right)\\ &=0 \end{aligned}

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##### Quiz-2 / TUT0402 quiz2
« on: October 04, 2019, 02:00:02 PM »
Determine whether the equation is exact, if it is exact find the solution
$$(2xy^2+2y)+(2x^2y+2x)y^{\prime}=0$$

$$\begin{array} { l } {M_y=4xy+2\quad N_x=4xy+2}\\ {M_y=N_x\rightarrow\text{exact}}\\ {G \varphi(x,y)\quad\text{such that}\quad \varphi(x)=M\quad\varphi(y)=N} \end{array}$$
$$\left. \begin{array} { l } { \varphi x = 2 x y ^ { 2 } + 2 y } \\ { \varphi = x ^ { 2 } y ^ { 2 } + 2 x y + h ( y ) } \\ { \varphi y = 2 x ^ { 2 } y + 2 x + h ^ { \prime } ( y ) = 2 x ^ { 2 } y + 2 x } \end{array} \right.$$

$$\left. \begin{array} { l } {h^{\prime}(y)=0}\\{h(y)=\text{constant}=c}\\ {x^{2} y ^ { 2 } + 2 x y = c } \end{array} \right.$$

7
##### Quiz-1 / TUT0402 Quiz1
« on: September 27, 2019, 02:01:57 PM »
Show that the given equation is homogenous and solve the differential equation:
$$\frac{d y}{d x}=-\frac{(4 x+3 y)}{(2 x+y)}$$
since the equation can be written as a ratio of y/x, it is homogenous
$$\frac{d y}{d x}=-\frac{(4 x+3 y)}{(2 x+y)}=-\frac{\left(4+3 \frac{y}{x}\right)}{\left(2+\frac{y}{x}\right)}$$
$$u=\frac{y}{x} \quad y=u x$$
$$\frac{d y}{d x}=\frac{d u}{d x} \cdot x+u=-\frac{(4+3 u)}{2+u}$$
$$\frac{\partial u}{d x} \cdot x=\frac{-4-3 u-2 u-u^{2}}{2+u}=\frac{-4-5 u-u^{2}}{2+u}$$
$$\begin{array}{l}{\frac{2+u}{u^{2}+5 u+4} \partial u=-\frac{1}{x} d x} \\ {\frac{2+u}{(u+1)(u+4)} \quad \partial u=-\frac{1}{x} d x}\end{array}$$
$$\begin{array}{l}{\frac{1}{3} \int \frac{1}{u+1} d u+\frac{2}{3} \int \frac{1}{u+4} d u=-\frac{1}{x} d x} \\ {\frac{1}{3} \ln |u+1|+\frac{2}{3} \ln |u+4|=-\ln |x|+C}\end{array}$$
$$\frac{1}{3} \ln \left|\frac{y+x}{x}\right|+\frac{2}{3} \ln \left|\frac{y+4 x}{x}\right|=-\ln |x|+C$$
$$\frac{1}{3} \ln |y+x|-\frac{1}{3} \ln |x|+\frac{2}{3} \ln |y+4 x|-\frac{2}{3} \ln |x|=-\ln |x|+c$$
$$\begin{array}{c}{\ln |y+x|+2 \ln |y+4 x|=36} \\ {\ln |y+x|y+4 x|^{2}=36} \\ {|y+x||y+4 x|^{2}=c}\end{array}$$

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