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### Messages - Yichen Ji

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##### Quiz-5 / TUT 0602 Quiz 5
« on: November 05, 2019, 03:12:52 AM »
Question: find the general solution of the differential equation $4y''+y=2sec(\frac{t}{2})$ where $-\pi<t<\pi$

Solution:
By the method of variation of parameters, first we find the solution of the homogeneous equation
\begin{equation*}
4y''+y=0
\end{equation*}
The characteristic polynomial equation
\begin{equation*}
4r^2+1=0
\end{equation*}
We get $r_1=\frac{1}{2}i$ and $r_2=-\frac{1}{2}i$
Then the homogeneous solution
\begin{equation*}
y_h(t)=c_1cos(\frac{t}{2})+c_2sin(\frac{t}{2})
\end{equation*}
Then, substitute the two constants$c_1$ and $c_2$ for $u_1$ and $u_2$, we get

y=u_1cos(\frac{t}{2})+u_2sin(\frac{t}{2})

Take differentiation, we get
\begin{equation*}
y'=u_1'cos(\frac{t}{2})+u_2'sin(\frac{t}{2})-\frac{1}{2}u_1sin(\frac{t}{2})+\frac{1}{2}u_2cos(\frac{t}{2})
\end{equation*}
Set

u_1'cos(\frac{t}{2})+u_2'sin(\frac{t}{2})=0

So
\begin{equation*}
y'=-\frac{1}{2}u_1sin(\frac{t}{2})+\frac{1}{2}u_2cos(\frac{t}{2})
\end{equation*}
Differentiate again,we get

y''=-\frac{1}{2}u_1'sin(\frac{t}{2})+\frac{1}{2}u_2'cos(\frac{t}{2})-\frac{1}{4}u_1cos(\frac{t}{2})-\frac{1}{4}u_2sin(\frac{t}{2})

Substitute (1)(3) back to the original equation and use (2) for substitution,we get
\begin{equation*}
u_2'\frac{sin(\frac{t}{2})}{cos(\frac{t}{2})}sin(\frac{t}{2})+u_2'cos(\frac{t}{2})=\frac{1}{cos(\frac{t}{2})}
\end{equation*}
Reorganizing, we get
\begin{align*}
u_2' &=1\\
u_2 &=t+c_2\\
u_1' &=-tan(\frac{t}{2})\\
u_1 &=2ln|cos(\frac{t}{2})|+c_1
\end{align*}
Finally we get the general solution
\begin{equation*}
y=c_1cos(\frac{t}{2})+c_2sin(\frac{t}{2})+2ln|cos(\frac{t}{2})|cos(\frac{t}{2})+tsin(\frac{t}{2})
\end{equation*}
where$-\pi<t<\pi$

2
##### Quiz-2 / TUT 0602 Quiz 2
« on: November 04, 2019, 06:40:19 PM »
Question: show that the given equation is not exact but becomes exact when multiplied by the given integrating factor.Then solve the equation.
The equation is:
\begin{equation*}
x^2y^3+x(1+y^2)y'=0
\end{equation*}
The integrating factor $\mu(x,y)=\frac{1}{xy^3}$

Solution:
First,verify that the initial equation is not exact:
\begin{align*}
M(x,y) &=x^2y^3 & N(x,y) &=x+xy^2\\
M_y &=3x^2y^2 & N_x &=1+y^2\\
\end{align*}
$M_y \neq N_x$,the equation is not exact.
Next, multiplying $\mu$ on both sides,
\begin{align*}
\mu x^2y^3+\mu x(1+y^2)y'=0\\
x+\frac{1+y^2}{y^3}y'=0
\end{align*}
Denote $P(x,y)=x$ and $Q(x,y)=\frac{1+y^2}{y^3}$
Check exactness:
\begin{equation*}
P_y=Q_x=0
\end{equation*}
The new equation is exact.
Now solve the equation:
\begin{align*}
\frac{\partial\psi}{\partial x}=x\\
\psi=\frac{1}{2}x^2+g(y)\\
\frac{\partial\psi}{\partial y}=g'(y)=y^{-3}+y^{-1}\\
g(y)=-\frac{1}{2}y^{-2}+ln|y|
\end{align*}
So the solution to the equation is
\begin{equation*}
\psi(x,y)=\frac{x^2}{2}-\frac{1}{2y^2}+ln|y|=c
\end{equation*}
for $c$ being constant.

3
##### Quiz-1 / TUT 0602 Quiz 1
« on: November 04, 2019, 05:58:57 PM »
Question:Solve the given differential equation:
\begin{equation*}
\frac{dy}{dx}=\frac{x-e^{-x}}{y+e^{y}}
\end{equation*}
Solution:
Reordering:
\begin{equation*}
(y+e^{y})dy=(x-e^{-x})dx
\end{equation*}
Take integration and multiplying$2$on both sides:
\begin{equation*}
y^2+2e^{y}=x^2+2e^{-x}+c
\end{equation*}
for $c$ being constant.
Reordering,get implicit solution
\begin{equation*}
y^2-x^2+2e^{y}-2e{-x}=c
\end{equation*}

4
##### Quiz-3 / TUT 0602 Quiz 3
« on: November 04, 2019, 05:36:53 PM »
Question:find a differential equation whose solution is $y=c_1e^{2t}+c_2e^{-3t}$
Solution:
Set
\begin{align*}
y_1 &=e^{2t} & y_2 &=e^{-3t}\\
y_1' &=2e^{2t} & y_2' &=-3e^{-3t}\\
r_1 &=2 & r_2 &=-3\\
\end{align*}
So the Wronskian
\begin{equation*}
\begin{split}
W[y_1,y_2](t) &= \begin{vmatrix}
y_1(t)&y_2(t)\\
y_1'(t)&y_2'(t)
\end{vmatrix} \\
&=y_1y_2'-y_2y_1' \\
&=-5e^{-t}\\
&\neq0
\end{split}
\end{equation*}
Then ${y_1,y_2}$forms a fundamental set of solution.We can write one characteristic equation as
\begin{align*}
(r-2)(r+3)=0\\
(r-2)(r+3)e^{rt}=0
\end{align*}
by multiplying $e^{rt}$on both sides,we get one equation for this general solution:
\begin{equation*}
y''+y'-6y=0
\end{equation*}

5
##### Quiz-4 / TUT 0602 Quiz 4
« on: November 04, 2019, 05:01:07 PM »
Question:find the general solution of the equation:

y''+y'-6y=12e^{3t}+12e^{-2t}

Solution:
First consider homogeneous equation $y''+y'-6y=0$
the characteristic polynomial equation is
\begin{align*}
r^2+r-6 &=0\\
(r+3)(r-2) &=0\\
\end{align*}
We get $r_1=-3$ and $r_2=2$
Therefore, the homogeneous solution is $y_h(t)=c_1e^{-3t}+c_2e^{2t}$
Next,we look for $y_p(t)$
Consider
\begin{align*}
y_1(t) &=Ae^{3t}\\
y_1'(t) &=3Ae^{3t}\\
y_1''(t) &=9Ae^{3t}\\
6Ae^{3t} &=12e^{3t}\\
A &= 2
\end{align*}
So $y_1(t)=2e^{3t}$
Then consider
\begin{align*}
y_2(t) &=Be^{-2t}\\
y_2'(t) &=-2Be^{-2t}\\
y_2''(t) &=4Be^{-2t}\\
-25Be^{-2t} &=12e^{-2t}\\
B &=-\frac{12}{25}\\
\end{align*}
So $y_2(t)=-\frac{12}{25}e^{-2t}$
And the particular solution is
\begin{align*}
y_p(t) &=y_1(t)+y_2(t)\\
y_p(t) &=2e^{3t}-\frac{12}{25}e^{-2t}\\
\end{align*}
To conclude, the general solution is
\begin{align*}
y_g(t) &=y_h(t)+y_p(t)\\
y_g(t) &=c_1e^{-3t}+c_2e^{2t}+2e^{3t}-\frac{12}{25}e^{-2t}\\
\end{align*}

6
##### Quiz-4 / Re: TUT0801
« on: October 19, 2019, 03:38:57 PM »
Which section does this question belong to?

7
##### Quiz-4 / Re: QUIZ4 TUT 0502
« on: October 19, 2019, 03:32:50 PM »
How did you determine the partial derivative of which variable(x/y) of μto be 0? Did you calculate both and choose the easier one?

8
##### Quiz-1 / Re: TUT0303 Quiz 1
« on: October 02, 2019, 09:50:22 PM »
A quick question: We can directly observe that this function is homogeneous(highest power = 2). Do we need to show that or can we use substitution without showing that?

9
##### Quiz-1 / Re: TUT0301
« on: October 02, 2019, 09:46:24 PM »
A typo reminder: the power of the first 2 terms should be 2t.

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