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Topics - AllanLi

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1
Quiz-4 / quiz 4 tut 0401
« on: October 18, 2019, 02:29:22 PM »
find the general solution to the given equation
\begin{equation}
9y''+6y'+y=0
\end{equation}we get
\begin{equation}
9r^2+6r+1=0
\end{equation}solve for r, we have
\begin{equation}
3r+1=0, r = -1/3
\end{equation} so we have
\begin{equation}
y(t) = C1e^{-1/3}+C2te^{-1/3}
\end{equation}

2
Quiz-3 / quiz 3 tut 0401
« on: October 11, 2019, 02:02:30 PM »
\begin{equation}
y'' + 3y' = 0 , y(0) = -2, y'(0) = 3
\end{equation}we get
\begin{equation}
r^2 + 3r = 0
\end{equation} Solve for r, we get the solution for r.
\begin{equation}
r1= -3, r2 = -3
\end{equation} For the repeated roots, the solution for y is
\begin{equation}
y(t) = C1e^{rt} + C2te^{rt}
\end{equation}So we have
\begin{equation}
y(t) = C1e^{-3t}+tC2e^{-3t}
\end{equation} Since
\begin{equation}
y(0) = -2, y'(0) = 3
\end{equation} We will have two equations about C1 and C2
\begin{equation}
C1 = -2, -3C1 + C2(1+0) = 3
\end{equation}So we have
\begin{equation}
C1 = -2, C2 = -3
\end{equation} Then the solution is
\begin{equation}
y(t) = -2e^{-3t}-3te^{-3t}
\end{equation}
\begin{equation}
y(t) = - (2+3t)e^{-3t}
\end{equation}

3
Quiz-2 / quiz 2 tut 0401
« on: October 04, 2019, 02:00:02 PM »
\begin{equation}
(2xy^2+2y)+(2x^2y+2x)y'=0
\end{equation} write y' into dy/dx, then we get
\begin{equation}
(2xy^2+2y)dx+(2x^2y+2x)dy=0
\end{equation} Let M = 2xy^2+2y , N = 2x^2y+2x , we get
\begin{equation}
\frac{dM}{dy}= 4xy + 2 , \frac{dN}{dx} = 4xy + 2
\end{equation} Since dM/dy = dN/dx, therefore they are exact. So ∃ 𝛗(x,y) = ∫ M dx
\begin{equation}
𝛗(x,y) = ∫ M dx = ∫ 2xy^2 + 2y dx = x^2y^2 + 2xy + h(y)
\end{equation}d𝛗(x,y)/dy = N
\begin{equation}
4x^2y + 2x +h'(y) = 2x^2y + 2x
\end{equation} we get h'(y) = 0 => h(y) = C, C is a constant.At the end we got the answer
\begin{equation}
𝛗(x,y) = x^2y^2 + 2xy +C
\end{equation}

4
Quiz-1 / quiz 1 tut 0401
« on: September 27, 2019, 02:01:31 PM »
       xy'=(1-y^2)^(1/2)
 write y' into dy/dx then we get
  x * (dy/dx) = (1-y^2)^(1/2)
 Then we rearrange the equation, let x,dx and y,dy on the same side.
  we get 1/x * dx = (1-y^2)^(-1/2) * dy
 by taking integral on both side,
    ln|x|+C = arcsin(y)
therefore y = sin(ln|x|+C)

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