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### Messages - lilywq

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1
##### Chapter 1 / How to solve the question on section 1.4 of the text book Page42
« on: October 01, 2020, 09:02:38 PM »
On the page 42 of the text book. Here is the question:

Find the limit at ∞ of the given function, or explain why it does not exist.
24. h(z) = Arg(z), z≠0

I wonder is the limit do not exist? and why this limit do not exist?

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##### Quiz-4 / Quiz-4 TUT0102
« on: October 18, 2019, 09:22:41 PM »
\begin{align*}
y^"+y &=  3sin(2t)+tcos(st)
\end{align*}
First consider the homogeneous differential equation for finding complimentary solution
\begin{align*}y^"+y &= 0
\end{align*}
{Assume} y=e$^{rx}$ \text {be the solution of the equation,then}
\begin{align*}r^2+1&=0\\
r^2&=-1\\
r &=\pm i\end{align*}
Therefore, the roots are r $=\pm i$\\
Therefore, the complementary solution of the given differential equation is
\begin{align*}
y_c(t)&=c_{1}cos(t)+c_{2}sin(t)\end{align*}
the particular solution for the given differential equation is of the following form:
y$_{p}$ = (At+B)cos(2t)+(Ct+D)sin(2t)\\
\text {Differentiate} y$_{p}$ \text {with respect to t as follows:}\\
\begin{align*}
y^{'}_{p} &= -2(At+B)sin(2t)+Acos(2t)+2(Ct+D)cos(2t)+Csin(2t)\\
y^{'}_{p}&=(-2At-2B+C)sin(2t)+(2Ct+2D+A)cos(2t)\\
y^{"}_{p}&=(-4At-4B+4C)cos(2t)-(4Ct+4D+4A)sin(2t)\\
\end{align*}
Substitue y$_{p}$ and y$^{"}_{p}$ in equation\\
\begin{align*}y^{"}+y^{'} &= 3sin2t + t cos2t\\
-2(At+B)sin(2t)+Acos(2t)+2(Ct+D)cos(2t)+Csin(2t)+-2(At+B)sin(2t)+Acos(2t)+2(Ct+D)cos(2t)+Csin(2t) = 3sinwt + t cos2t\\
-Atcos(2t)+(-3B+4C)cos(2t)-3Ctsin(2t)+(-3D-4A)sin(2t) &= 3sin2t+tcos2t\\\end{align*}
Compare the coefficients of  tcos(2t) on both sides\\
\begin{align*}-3A&=1\\
A=-\frac{1}{3}\\\end{align*}
Compare the coefficients of  sin(2t) on both sides\\
\begin{align*}-3D-4A&=3\\
-3D-4(-\frac{1}{3})&=3\\
D&=-\frac{5}{9}\end{align*}
Compare the coefficients of  tsin(2t) on both sides\\
\begin{align*}-3C&= 0\\
C&=0
\end{align*}

3
##### Quiz-3 / TUT0102 Quiz3
« on: October 11, 2019, 03:02:25 PM »
\begin{align*}
y^"+8y^{'}-9y=0 && {y(1)=1,y'(1)=0}\\
r^2+8r-9&=0\\
(r+9)(r-1)&=0\\
r=-9 \text{ or } r=1\\
y&=c_{1}e^{-9t}+c_{2}e^t\\
y(1)&=c_1e^{-9(1)}+c_{2}e^{1}\\
1&=c_{1}e^{-9t}+c_{2}e^t\\
\text{Differentiate y with respect to t, we get}\\
y^{'}&= -9c_{1}e^{-9t}+c_{2}e^{t}\\
y^{'}(1)&= -9c_{1}e^{-9(1)}+c_{2}e^{1}\\
0 &= -9c_{1}e^{-9}+c_{2}e\\
9c_{1}e^{-9}&=c_{2}e\\
c_{1}&=\frac{c_{2}e^{10}}{9}\\
\text{Substitute } c_{1}=\frac{c_{2}e^{10}}{9} \text{ in } 1=c_{1}e^{-9}+c_{2}e\\
1&=(\frac{c_{2}e^{10}}{9})e^{-9}+c_{2}e\\
1&=\frac{10}{9}c_{2}e\\
c_{2}&=\frac{9}{10e}\\
c_{1}&=\frac{1}{10}e^9\\
\text{Therefore, the general solution of the initial value problem(1) is}\\
y=\frac{1}{10}e^{9(1-t)}+\frac{9}{10}e^{t-1}
\end{align*}

4
##### Quiz-2 / Quiz-2 / TUT0102
« on: October 04, 2019, 04:44:01 PM »
\begin{align*}
x^2y^3 + x(1+y^2)y' &= 0   && \equation{\mu = 1/xy^3}\\
1/xy^3*(x^2y^3 + x(1+y^2)y')  &= 0 \\
x+y^{-3}(1+y^2)y' &= 0\\
M=x,N=y^{-3}(1+y^2)\\
M_y(x,y)=0=N_x(x,y)\\
\psi_x(x,y)=M(x,y)=x, \psi_y(x,y)=N(x,y)=y^{-3}(1+y^2)\\
\psi=x^2/2+h(y)\\
\psi_y=h'(y)=y^{-3}(1+y^2)\\
h(y)=ln(x)-1/(2y^2)\\
\psi=x^2/2+ln(x)-1/(2y^2)=c\\
\psi=x^2+2ln(x)-y^{-2}=c
\end{align*}

5
##### Quiz-2 / Quiz-2 / TUT0102
« on: October 04, 2019, 04:41:26 PM »
\begin{align*}
x^2y^3 + x(1+y^2)y' &= 0   && \equation{\mu = 1/xy^3}\\
1/xy^3*(x^2y^3 + x(1+y^2)y')  &= 0 \\
x+y^{-3}(1+y^2)y' &= 0\\
M=x,N=y^{-3}(1+y^2)\\
M_y(x,y)=0=N_x(x,y)\\
\psi_x(x,y)=M(x,y)=x, \psi_y(x,y)=N(x,y)=y^{-3}(1+y^2)\\
\psi=x^2/2+h(y)\\
\psi_y=h'(y)=y^{-3}(1+y^2)\\
h(y)=ln(x)-1/(2y^2)\\
\psi=x^2/2+ln(x)-1/(2y^2)=c\\
\psi=x^2+2ln(x)-y^{-2})=c
\end{align*}

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