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Messages - AllanLi

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1
Term Test 1 / Re: Problem 1 (morning)
« on: October 23, 2019, 09:24:54 AM »
\begin{equation}
-ysinx+y^3cosx+(3cosx+5y^2sinx)y'=0
\end{equation}
\begin{equation}
(-ysinx+y^3cosx)dx+(3cosx+5y^2sinx)dy=0
\end{equation}let M = -ysinx+y^3cosx, and N = 3cosx+5y^2sinx
\begin{equation}
M_y = -sinx+3y^2cosx, N_x = -3sinx+5y^2cosx
\end{equation} let My-Nx
\begin{equation}
M_y - N_x = 2sinx-2y^2cosx
\end{equation}since this looks famillier with M , so we are taking R1
\begin{equation}
R_1 = \frac{-M_y+N_x}{M}, R_1 = \frac{-2sinx+2y^2cosx}{-ysinx+y^3cosx},R_1 = \frac{2}{y}
\end{equation} the the integrating factor 𝝻 will be the e to the power of integral of R1
\begin{equation}
𝝻 = e^{∫R_1dy},𝝻 = y^2
\end{equation}then we times y^2 in this equation ,we get
\begin{equation}
(-y^3sinx+y^5cosx)dx+(3y^2cosx+5y^4sinx)dy=0
\end{equation}therefore
\begin{equation}
𝛗 = ∫-y^3sinx+y^5cosxdx=y^3cosx+y^5sinx+h(y)
\end{equation} take the derivative on 𝛗 of y,we get
\begin{equation}
𝛗_y = 3y^2cosx+5y^4sinx+h'(y)=3y^2cosx+5y^4sinx
\end{equation}we will have h'(y) = 0, so h(y)=C is a constant
\begin{equation}
𝛗=y^3cosx+y^5sinx+C
\end{equation}
\begin{equation}
y^3cosx+y^5sinx+C=0
\end{equation}since we have
\begin{equation}
y(\frac{\pi_1}{4})=√2
\end{equation}solve for C
\begin{equation}
C =6
\end{equation}
\begin{equation}
y^3cosx+y^5sinx+6=0
\end{equation}

2
Term Test 1 / Re: Problem 3 (morning)
« on: October 23, 2019, 09:20:39 AM »
\begin{equation}
y''-6y'+8y=48sinh(2x)
\end{equation}applying the hint
\begin{equation}
y''- 6y' + 8y= 24e^{2x}-24e^{-2x}
\end{equation}solving the homogeneous equation
\begin{equation}
r^2-6r+8=0 ,  r_1 = 2, r_2 = 4
\end{equation}
\begin{equation}
Yc = C_1e^2 + C_2e^4
\end{equation}let Yp = Axe^{2x}
\begin{equation}
Yp' = A(e^{2x} + 2xe^{2x}), Yp'' = A(2e^2x + 2(e^{2x} +2xe^{2x}))
\end{equation} fill in the original equation
\begin{equation}
4Ae^{2x}+4Axe^{2x} - 6A(e^{2x}+2xe^{2x})+8Axe^{2x}=24e^{2x}
\end{equation}solve for A
\begin{equation}
A = -12
\end{equation} let
\begin{equation}
Ys = Be^{-2x}
\end{equation}
\begin{equation}
Ys' = -2Be^{-2x} , Ys'' = 4Be^{-2x}
\end{equation}
\begin{equation}
4Be^{-2x} -6(-2Be^{-2x}) + 8Be^{-2x} = -24e^{-2x}
\end{equation}then we have
\begin{equation}
24B = -24, B = -1
\end{equation} so we have general solution
\begin{equation}
y = c1e^{2x} + c2e^{4x} -12xe^{2x}-e^{-2x}
\end{equation}Part(b)
\begin{equation}
y(0)=0, y'(0)=0
\end{equation} we get
\begin{equation}
C_1+C_2-1 =0 , 2C_1 + 4C_2 -12+2 = 0
\end{equation}solve for C1 and C2
\begin{equation}
C_1 = -3 , C_2 = 4
\end{equation}Therefore the answer would be
\begin{equation}
y(x) = -3e^{2x}+4e^{4x}-12xe^{2x}-e^{-2x}
\end{equation}

3
Term Test 1 / Re: Problem 2 (morning)
« on: October 23, 2019, 09:14:42 AM »
\begin{equation}
xy''-(2x+1)y'+(x+1)y=0
\end{equation}
\begin{equation}
y''-\frac{2x+1}{x}y'+\frac{x+1}{x}y=0
\end{equation}p(x)=(2x+1)/x
\begin{equation}
𝝻=e^{-∫-\frac{2x+1}{x}dx}
\end{equation}W=C𝝻
\begin{equation}
𝝻=Ce^{2x}x
\end{equation}W=C𝝻
\begin{equation}
W = Ce^{2x}x
\end{equation}let C =1,then we have
\begin{equation}
W{(y_1,y_2)}=xe^{2x}
\end{equation}for part b), to check that y1 is a solution then
\begin{equation}
y_1' = e^x,y_1'' = e^x
\end{equation}
\begin{equation}
xe^x-(2x+1)e^x+(x+1)e^x=0
\end{equation}so y1 is indeed a solution.
\begin{equation}
W_{(y_1,y_2)}=xe^{2x}=e^xy_2'-e^xy_2
\end{equation}
\begin{equation}
y_2'-y_2=xe^x
\end{equation}the integrating factor is
\begin{equation}
𝝻 = e^{-x}
\end{equation} multiply 𝝻 in this equation
\begin{equation}
e^{-x}y_2'-e^{-x}y_2=x
\end{equation}
\begin{equation}
(e^{-x}y_2)'=x
\end{equation}take integral on both side
\begin{equation}
e^{-x}y_2=\frac{1}{2}x^2
\end{equation}move all the x to one side
\begin{equation}
y_2 = \frac{1}{2}x^2e^x
\end{equation}part c) , the general solution is
\begin{equation}
W = C_1y_1+C_2y_2
\end{equation}
\begin{equation}
y = C_1e^x+\frac{C_2}{2}x^2e^x
\end{equation}
\begin{equation}
y' = C_1e^x+\frac{C_2}{2}(2xe^x+x^2e^x)
\end{equation}we have y(1) = 0 and y(1)'=e
\begin{equation}
C_1e+\frac{C_2}{2}e=0, C_1e+\frac{C_2}{2}(2e+e)=e
\end{equation}solve for C1 and C2,then we have
\begin{equation}
C_1 =-\frac{1}{2},C_2 = 1
\end{equation}
\begin{equation}
y=-\frac{1}{2}e^x+\frac{1}{2}x^2e^x
\end{equation}

4
Term Test 1 / Re: Problem 4 (morning)
« on: October 23, 2019, 08:36:55 AM »
\begin{equation}
y’’- 6y'+25y=102sinx+16e^{3x}
\end{equation}first solving the homogenous equation
\begin{equation}
r^2-6r+25=0
\end{equation}solve for r, we get
\begin{equation}
r = 3 ± 4i
\end{equation} so we have 𝞴=3 and 𝝻=4. Then we have
\begin{equation}
y_c=c_1e^{3x}cos4x+c_2e^{3x}sin4x
\end{equation}let
\begin{equation}
y_p = asinx+bcosx,y_p'=acosx-bsinx,y_p'' = -asinx-bcosx
\end{equation}so we will have
\begin{equation}
-asinx-bcosx-6(acosx-bsinx)+25(asinx+bcosx)=102sinx
\end{equation}solve for a and b
\begin{equation}
a=4b,102b = 102
\end{equation}so a =4 , b = 1
\begin{equation}
y_{p2} = Ce^{3x},y_p2'=3Ce^{3x},y_p2'' = 9Ce^{3x}
\end{equation}
\begin{equation}
9C-18C+25C=16
\end{equation}so we have C=1
\begin{equation}
y_{p2} = e^{3x}
\end{equation}
\begin{equation}
y= c_1e^{3x}cos4x+c_2e^{3x}sin4x+e^{3x}+4sinx+cosx
\end{equation}next step, we are looking for the derivative of y
\begin{equation}
y'= c_1(3e^{3x}cos4x-4e^{3x}sin4x)+c_2(3e^{3x}sin4x+4e^{3x}cos4x)+3e^{3x}+4cosx-sinx
\end{equation}since we have y(0)=0, y'(0)=0, so we get 2 functions about the coefficients. By the way, cos0=1 and sin0 =0.
\begin{equation}
c_1=-2, 3c_1+4c_2+7=0
\end{equation}
\begin{equation}
c_1 =-2 , c_2 = -1/4
\end{equation} the final answer will be
\begin{equation}
y =-2e^{3x}cos4x-\frac{1}{4}e^{3x}sin4x+e^{3x}+4sinx+cosx
\end{equation}

5
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}

6
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}

7
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}

8
Quiz-1 / Re: quiz 1 tut 0401
« on: September 27, 2019, 05:05:09 PM »

9
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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