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### Messages - Meng Wu

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46
##### Quiz-4 / Re: Q4-T0501
« on: March 02, 2018, 04:58:00 PM »
$$(1-t)y’’+ty’-y=2(t-1)^2e^{-t}, 0< t<1; y_1(t)=e^t, y_2(t)=t$$
Hence,$$\cases{y_1(t)=e^t\\y_1’(t)=e^t\\y_1’’(t)=e^t} \text{and} \cases{y_2(t)=t\\y_2’(t)=1\\y_2’’(t)=0}$$
Substitute back into the homogeneous equation: $$(1-t)y’’+ty’-y=0$$
Verified that $y_1(t)$ and $y_2(t)$ both satisfy the corresponding homogeneous equation. $\\$
And the complementary solution $y_c(t)=c_1e^t+c_2$ $\\$
Now divide both sides of the original equation by $1-t$:
$$y’’+{t\over 1-t}-{1\over 1-t}=-2(t-1)e^{-t}$$
Then $$p(t)={t\over 1-t}, q(t)=-{1\over 1-t}, g(t)=-2(t-1)e^{-t}$$
$$W[y_1,y_2](t)=\begin{array}{|c c|} y_1(t)&y_2(t)\\y_1’(t)&y_2’(t)\end{array}=(1-t)e^t$$
Since the particular solution has the form: $$Y(t)=u_1(t)y_1(t)+u_2(t)y_2(t)$$
and \begin{align}u_1(t)&=-\int{{y_2(t)g(t)\over W[y_1,y_2](t)}}dt\\&=-\int{t\cdot (-2(t-1)e^{-t})\over (1-t)e^t}dt\\&=-2\int{te^{-2t}}dt\\&=(t+{1\over2})e^{-2t}\end{align}
\begin{align}u_2(t)&=\int{y_1(t)g(t)\over W[y_1,y_2](t)}dt\\&=\int{e^t\cdot (-2(t-1)e^{-t})\over (1-t)e^t}dt\\&=2\int{e^{-t}}\\&=-2e^{-t}\end{align}
Therefore, $$Y(t)=(t+{1\over2})e^{-2t}\cdot e^t+ (-2e^{-t})\cdot t=({1\over2}-t)e^{-t}$$
Hence, the general solution:
\begin{align}y(t)&=y_c(t)+Y(t)\\&=c_1e^t+c_2t+({1\over2}-t)e^{-t}\end{align}
Therefore, the particular solution of the given nonhomogeneous equation is $$Y(t)=({1\over2}-t)e^{-t}$$

47
##### Quiz-4 / Q4-T0501
« on: March 02, 2018, 04:56:49 PM »
Verify the given functions $y_1$ and $y_2$ satisfy the correspending homogeneous equation; then find a particular solution of the given nonhomogeneous equation.
\begin{align*}
&(1-t)y’’+ty’-y=2(t-1)^2e^{-t}, 0< t<1\\
&y_1(t)=e^t, y_2(t)=t.
\end{align*}

48
##### Term Test 1 / Re: P3-Morning
« on: February 21, 2018, 10:56:39 AM »
Small Error:
For part(b), $y'(0)$ should be $$y'(0)=3c_1+2c_2+2=0$$
$$\implies\cases{c_1=2\\c_2=-4}$$ and $$y(t)=2e^{3t}-4e^{2t}+2e^t-te^{2t}$$

49
##### Term Test 1 / Re: P2-Morning
« on: February 21, 2018, 10:28:50 AM »
Cheryl

Meng
(9) is OK, but (10) is not

50
##### Term Test 1 / Re: P1-Day
« on: February 21, 2018, 08:31:10 AM »
(4) is correct but then integrated with an error

Typos. Corrections done.

51
##### Term Test 1 / Re: P2-Morning
« on: February 16, 2018, 01:15:55 AM »
$(a)$ $\\$
First we divide both sides by $-x^2(ln(x)-1)$:
$$y’’-{x \over x^2(ln(x)-1)}y’+{1 \over x^2(ln(x)-1)}y=0$$
Let $p(t)=-{x \over x^2(ln(x)-1)}$ and $q(t)={1 \over x^2(ln(x)-1)}.$ $\\$
Noting $p(t)$ and $q(t)$ are continuous everywhere except when $x^2(ln(x)-1)\neq0$.
$\\$
By Abel’s Theorem:\begin{align}W(y_1,y_2)(x)&=cexp(-\int{p(x)dx})\\&=cexp(-\int-{x \over x^2(ln(x)-1)}dx)\\&=ce^{ln|ln|x|-1|}\\&=c(ln|x|-1)\end{align}
Let $c=1$, then $W(y_1,y_2)(x)=ln|x|-1$. $\\$
$(b)$ $\\$
Since $y_1(x)=x,$ then
$$\cases{y_1’(x)=1\\y_1’’(x)=0}$$
Substitute these values back to the ODE:
\begin{align}0-{x \over x^2(ln(x)-1)}\cdot 1 + {1 \over x^2(ln(x)-1)}\cdot x=0\end{align}
Thus $y_1=(x)$ is indeed a solution. $\\$
Now we need to find $y_2(x)$ $\\$
Since we already know: \begin{align}W(y_1,y_2)(x)&=\begin{array}{|c c|}y_1(x)& y_2(x) \\ y_1’(x) & y_2’(x)\end{array}=ln|x|-1\\&=\begin{array}{|c c|}x& y_2(x) \\ 1& y_2’(x)\end{array}=xy_2’(x)-y_2(x)=ln|x|-1\end{align}
Hence, we have $$xy_2’-y_2=ln|x|-1$$
Divide both sides by $x$:
$$y_2’-{1\over x}y_2={ln|x|\over x}-{1 \over x}$$
$$\mu(x)=exp(\int-{1\over x}dx)=e^{-ln|x|}=x^{-1}$$
Multiply both sides by $x^{-1}:$ $\\$
\begin{align}x^{-1}y_2’-x^{-2}y_2&=x^{-2}ln|x|-x^{-2}\end{align}
Hence, $$(x^{-1}y_2)’=x^{-2}ln|x|-x^{-2}$$
Integrating both sides:
$$\int{(x^{-1}y_2)’}dx=\int{x^{-2}\ln|x|-x^{-2}}dx$$
\begin{align}x^{-1}y_2&=\int{x^{-2}\ln|x|}dx-\int{x^{-2}}dx\end{align}
For $\int{x^{-2}\ln|x|}dx$, we use Integral by parts: $\\$
Let $\cases{u=\ln|x|\\du={1\over x}dx}$   and $\cases{dv=x^{-2}dx\\v=-x^{-1}}$ $\\$
Thus:\begin{align}\int{x^{-2}\ln|x|}dx&=-x^{-1}\ln|x|-\int{-x^{-1}\cdot {1\over x}}dx\\&=-x^{-1}\ln|x|+\int{x^{-2}}dx \\&= -x^{-1}\ln|x|-x^{-1}+c\end{align}
Therefore,\begin{align}x^{-1}y_2&=\int{x^{-2}\ln|x|}dx-\int{x^{-2}}dx\\&=-x^{-1}\ln|x|-x^{-1}+c-\int{x^{-2}}dx\\&=-x^{-1}\ln|x|-x^{-1}+c+x^{-1}\\&=-x^{-1}\ln|x|+c\end{align}
Hence,$$y_2=-\ln|x|+cx$$
Let $c=1$, we have $$y_2=-\ln|x|+x$$
Since we already know that $W(y_1,y_2)(x)=\ln|x|-1\neq 0$,
Hence $y_2(x)$ is indeed another linearly independent solution. $\\$
$(c)$ $\\$
For $W(y_1,y_2)(x)=\ln|x|-1\neq 0$, we have general solution $$y=c_1y_1+c_2y_2=c_1x+c_2(-\ln|x|+x)$$ where $c_1$ and $c_2$ are some arbitrary constants.
Now, $$y’=c_1-{c_2\over x}+c_2$$
Let $x=1$ and $y=1$, $x=1$ and $y’=0$, we have
$$\cases{c_1+c_2=1\\c_1=0}\implies \cases{c_1=0\\c_2=1}$$
Therefore, the solution for the IVP is $$y(x)=-\ln|x|+x$$

52
##### Term Test 1 / Re: P1-Day
« on: February 15, 2018, 11:52:19 PM »
Let $$M(x,y)=2xy, N(x,y)=4x^2+4e^y+ye^y$$
Check exactness:
$${\partial{M} \over \partial{y}}=2x, {\partial{N} \over \partial{x}}=8x$$
Since, ${\partial{M} \over \partial{y}} \neq {\partial{N} \over \partial{x}}$, we need to find an integrating factor $\mu(x,y)$. $\\$
Now check if $\mu(x,y)$ if depends on only $x$, or depends on only $y$, or depends on only $xy$.
$${N_x-M_y\over M}={8x-2x\over 2xy}={3 \over y}$$
Thus the integrating factor $\mu(x,y)$ is depends on only $y$ which is $\mu(y)={ 3\over y}$. $\\$
Hence $${d\mu\over dy}={3\over y}\mu \implies \mu'-{3\over y}\mu=0$$
Integrating factor $\mu_1(y)=exp^{\int{p(y)}dy}=e^{3ln|y|}=y^3$ $\\$
Now we multiply $\mu_1(y)=y^3$ to both sides of the equation:
\begin{align}y^3\cdot 2xy+y^3\cdot(4x^2+4e^y+ye^y)&=0 \\ 2xy^4+(4x^2y^3+4y^3e^y+y^4e^y)y' &=0\end{align}
Now the differential equation becomes Exact:$\\$
So there is a function $\psi(x,y)$ such that:
\begin{align}\psi_x(x,y)&=2xy^4 \\ \psi_y(x,y)&=4x^2 y^3+4y^3e^y+y^4e^y\end{align}
We integrate $\psi_x(x,y)$ with respect to $x$:
$$\psi(x,y)=x^2y^4+h(y)$$
Hence,\begin{align}\psi_y(x,y)&=4x^2y^3+h'(y)\\&=4x^2y^3+4y^3e^y+y^4e^y\end{align} and
$$h'(y)=4y^3e^y+y^4e^y \implies h(y)=y^4 e^y$$
Therefore, $$\psi(x,y)=x^2y^4+y^4 e^y$$ and the general solution of this ODE is $$x^2y^4+y^4 e^y=c$$
Now let $x=1$ and $y=1$:
$$(1)^2 (1)^4+(1)^4 e^1=c$$
Therefore, the solution of the IVP is $$x^2y^4+y^4 e^y=e+1$$

53
##### Term Test 1 / Re: P4-Day
« on: February 15, 2018, 10:35:21 PM »
$(a)$ $\\$

First find the complementary solution for the homogeneous equation:
$$y’’-4y’+5y=0$$
Characteristic equation: $$r^2-4r+5=0 \implies \cases{r_1=2+i\\r_2=2-i}$$
Thus $$y_c(t)=c_1e^{2t}cos(t)+c_2e^{2t}sin(t)$$
Now we need to find the particular solution:
$$y_p(t)=Y_1(t)+Y_2(t)$$
We assume $Y_1(t)=Ae^t$ and there are no duplicates of $y_c(t)$,
$\\$thus $Y_1’(t)=Ae^t$ and $Y_1’’(t)=Ae^t$. $\\$
Substitue theses values back to $y’’-4y’+5y=2e^t$:
$$Ae^t-4Ae^t+5Ae^t=2e^t \implies A=1$$
We assume $Y_2(t)=Bcos(t)+Csin(t)$ and there are no duplicates of $y_c(t)$, $\\$thus $Y_2’(t)=-Bsin(t)+Ccos(t)$ and $Y_2’’(t)=-Bcost(t)-Csin(t)$. $\\$
Substitute these values back to $y’’-4y’+5y=8cos(t)$:
$$-Bcost(t)-Csin(t)+4Bsin(t)-4Ccos(t)+5Bcos(t)+5Csin(t)=8cos(t) \implies \cases{B=1\\C=-1}$$
Thus, $$y_p(t)=Y_1(t)+Y_2(t)=e^t+cos(t)-sin(t)$$
Therefore, the general solution is \begin{align}y(t)&=y_c(t)+y_p(t)\\&=c_1e^{2t}cos(t)+c_2e^{2t}sin(t)+e^t+cos(t)-sin(t)\end{align}

$(b)$ $\\$
$$y’(t)=2c_1e^{2t}cos(t)-c_1e^{2t}sin(t)+2c_2e^{2t}sin(t)+c_2e^{2t}cos(t)+e^t-sin(t)-cos(t)$$
Set $t=0$ and $y=0$; $t=0$ and $y’(t)=0$:
$$\cases{c_1+0+1+1-0=0\\2c_1-0+0+c_2+1-0-1=0} \implies \cases{c_1=-2\\c_2=4}$$
Therefore, the solution for the IVP is $$y(t)=-2e^{2t}cos(t)+4e^{2t}sin(t)+e^t+cos(t)-sin(t)$$

54
##### Term Test 1 / Re: P3-Day
« on: February 15, 2018, 09:58:07 PM »
$(a)$ $\\$
First we find the solution to the homogeneous solution for $$y’’(t)-3y’(t)+2y(t)=0$$
Characteristic equation: $$r^2-3r+2$$
Hence, $$\cases{r_1=1\\r_2=2}$$ and the complementary solution for the homogeneous equation is $$y_c(t)=c_1e^t+c_2e^{2t}$$
Now we need to find the particular solution for the non-homogeneous equation
$$y’’(t)-3y’(t)+2y(t)=-6+3e^t$$
Let $y_p(t)$ be the particular solution and $y_p(t)=Y_1(t)+Y_2(t)$.
We assume$Y_1(t)=At+B$, and there are no duplicates of the solution of the homogeneous equation, thus $Y_1’(t)=A$ and $Y_1’’(t)=0$. $\\$
And substitute these back to $y’’(t)-3y’(t)+2y(t)=-6$:
$$0-3A+2(At+B)=-6 \implies \cases{A=0\\B=-3}$$
We assume $Y_2(t)=Ce^t$, and noting there are duplicates of the solution of the homogeneous equation. $\\$ So we need to multiply $t$ for the propose of $Y_2(t)$, thus $Y_2(t)=Cte^t$ which leads to no duplicates. $\\$
Thus $Y_2’(t)=Ce^t+Cte^t$ and $Y_2’’(t)=2Ce^t+Cte^t$.
And substitute these values back to $y’’(t)-3y’(t)+2y(t)=3e^t$:
$$2Ce^t+Cte^t-3Ce^t-3Cte^t+2Cte^t=3e^t \implies C=-3$$
Hence, the particular solution $$y_p(t)=Y_1(t)+Y_2(t)=-3-3e^t$$
Therefore, the general solution is $$y(t)=y_c(t)+y_p(t)=c_1e^t+c_2e^{2t}-3-3te^t$$

$(b)$ $\\$
$$y’(t)=c_1e^t+2c_2e^{2t}-3e^t-3te^t$$
Set $t=0$ and $y=0$; $t=0$ and $y’=0$:
$$\cases{c_1+c_2=3\\c_1+2c_2=3} \implies \cases{c_1=3\\c_2=0}$$
Therefore, the solution for the IVP is $$y(t)=3e^t-3te^t-3$$

55
##### Term Test 1 / Re: P2-Day
« on: February 15, 2018, 09:16:18 PM »
$(a)$ $\\$
First we divide both sides by $x^2-1$:
$$y’’-{2x\over x^2-1}y’+{2 \over x^2-1}y=0$$
Let $p(t)=-{2x\over x^2-1}$ and $q(t)={2 \over x^2-1}.$ $\\$
Noting $p(t)$ and $q(t)$ are continuous everywhere except when $x=1$ or $x=-1.$
By Abel’s Theorem:\begin{align}W(y_1,y_2)(x)&=cexp(-\int{p(x)dx})\\&=cexp(-\int{-{2x\over x^2-1}}dx)\\&=ce^{ln|x^2-1|}\\&=c(x^2-1)\end{align}
Let $c=1$, then $W(y_1,y_2)(x)=x^2-1$. $\\$
$(b)$ $\\$
Since $y_1(x)=x,$ then
$$\cases{y_1’(x)=1\\y_1’’(x)=0}$$
Substitute these values back to the ODE:
\begin{align}0-{2x\over x^2-1}\cdot 1+{2 \over x^2-1}\cdot x=0\end{align}
$$-{2x\over x^2-1}+{2x\over x^2-1}=0$$
Thus $y_1=(x)$ is indeed a solution. $\\$
Now we need to find $y_2(x)$ $\\$
Since we already know: \begin{align}W(y_1,y_2)(x)&=\begin{array}{|c c|}y_1(x)& y_2(x) \\ y_1’(x) & y_2’(x)\end{array}=x^2-1\\&=\begin{array}{|c c|}x& y_2(x) \\ 1& y_2’(x)\end{array}=xy_2’(x)-y_2(x)=x^2-1\end{align}
Hence, we have $$xy_2’-y_2=x^2-1$$
Divide both sides by $x$:
$$y_2’-{1\over x}y_2=x-{1 \over x}$$
$$\mu(x)=exp(\int-{1\over x}dx)=e^{-ln|x|}=x^{-1}$$
Multiply both sides by $x^{-1}:$ $\\$
\begin{align}x^{-1}y_2’-x^{-2}y_2&=1-x^{-2}\end{align}
Hence, $$(x^{-1}y_2)’=1-x^{-2}$$
Integrating both sides:
$$\int{(x^{-1}y_2)’}dx=\int{1-x^{-2}}dx$$
\begin{align}x^{-1}y_2&=\int{1dx}-\int{x^{-2}}dx\\&=x+x^{-1}+c\end{align}
Hence,$$y_2=x^2+1+cx$$
Let $c=1$, we have $$y_2=x^2+x+1$$
Since we already know that $W(y_1,y_2)(x)=x^2-1\neq 0$,
Hence $y_2(x)$ is indeed another linearly independent solution. $\\$
$(c)$ $\\$
For $W(y_1,y_2)(x)=x^2-1\neq 0$, we have general solution $$y=c_1y_1+c_2y_2=c_1x+c_2(x^2+x+1)$$ where $c_1$ and $c_2$ are some arbitrary constants.
Now, $$y’=c_1+2c_2x+c_2$$
Let $x=0$ and $y=1$, $x=0$ and $y’=2$, we have
$$\cases{c_2=1\\c_1+c_2=2}\implies \cases{c_1=1\\c_2=1}$$
Therefore, the solution for the IVP is $$y(x)=x^2+2x+1$$

56
##### Term Test 1 / Re: P-2
« on: February 14, 2018, 10:22:08 AM »
Solution to Problem 2:

Prof. Victor would prefer you typing out the solutions xD ( that is if you want the bonus mark)

57
##### Quiz-3 / Re: Q3-T0301
« on: February 11, 2018, 10:06:57 AM »
First, we divide both sides of the equation by $cos(t)$:
$$y''+tan(t)y'-{t\over cos(t)}y=0$$
Now the given second-order differential equation has the form:
$$L[y]=y''+p(t)y'+q(t)y=0$$
Noting if we let $p(t)=tan(t)$ and $q(t)=-{t\over cos(t)}$, then $p(t)$ is continuous everywhere except at ${\pi\over 2}+k\pi$, where $k=0,1,2,\dots$ and $q(t)$ is also continuous everywhere except at $t=0$. $\\$
Therefore, by Abel's Theorem: the Wronskian $W[y_1,y_2](t)$ is given by
\begin{align}W[y_1,y_2](t)&= cexp(-\int{p(t)dt})\\&=cexp(-\int{tan(t)dt})\\&=ce^{ln|cos(t)|}\\&=ccos(t)\end{align}

58
##### Quiz-3 / Re: Q3-T0101
« on: February 11, 2018, 09:53:26 AM »
First, we divide both sides of the equation by $t^2$:
$$y''+{t+2\over t}y'+{t+2\over t^2}y=0$$
Now the given second-order differential equation has the form:
$$L[y]=y''+p(t)y'+q(t)y=0$$
Noting if we let $p(t)={t+2\over t}$ and $q(t)={t+2\over t^2}$, then $p(t)$ is continuous everywhere except at $t=0$, and $q(t)$ is also continuous everywhere except at $t=0$. $\\$
Therefore, by Abel's Theorem: the Wronskian $W[y_1,y_2](t)$ is given by
\begin{align}W[y_1,y_2](t)&= cexp(-\int{p(t)dt})\\&=cexp(-\int{{t+2\over t}dt})\\&=ce^{t+2ln|t|}\\&=ct^2e^t\end{align}

59
##### Quiz-3 / Re: Q3-T0201
« on: February 11, 2018, 09:34:16 AM »
$$y’’+3y’+2y=0$$
We assume that $y=e^{rt}$, and then it follows that $r$ must be a root of characteristic equation $$r^2+3r+2=(r+1)(r+2)=0$$
$$\cases{r_1=-1\\r_2=-2}$$
Since the general solution has the form of $$y=c_1e^{r_1t}+c_2e^{r_2t}$$
Therefore, the general solution of the given differential equation is
$$y=c_1e^{-t}+c_2e^{-2t}$$

60
##### Quiz-3 / Re: Q3-T0701
« on: February 11, 2018, 09:33:51 AM »
$$y''-2y'-2y=0$$
We assume that $y=e^{rt}$, and then it follows that $r$ must be a root of characteristic equation $$r^2-2r-2=0$$
We use the quadratic formula which is
$$r={-b\pm \sqrt{b^2-4ac}\over 2a}$$
Hence,
$$\cases{r_1={1+\sqrt{3}}\\r_2=1-\sqrt{3}}$$
Since the general solution has the form of $$y=c_1e^{r_1t}+c_2e^{r_2t}$$
Therefore, the general solution of the given differential equation is
$$y=c_1e^{(1+\sqrt{3})t}+c_2e^{(1-\sqrt{3})t}$$

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