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### Messages - AllanLi

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##### Term Test 1 / Re: Problem 1 (morning)
« on: October 23, 2019, 09:24:54 AM »

-ysinx+y^3cosx+(3cosx+5y^2sinx)y'=0

(-ysinx+y^3cosx)dx+(3cosx+5y^2sinx)dy=0
let M = -ysinx+y^3cosx, and N = 3cosx+5y^2sinx

M_y = -sinx+3y^2cosx, N_x = -3sinx+5y^2cosx
let My-Nx

M_y - N_x = 2sinx-2y^2cosx
since this looks famillier with M , so we are taking R1

R_1 = \frac{-M_y+N_x}{M}, R_1 = \frac{-2sinx+2y^2cosx}{-ysinx+y^3cosx},R_1 = \frac{2}{y}
the the integrating factor 𝝻 will be the e to the power of integral of R1

𝝻 = e^{∫R_1dy},𝝻 = y^2
then we times y^2 in this equation ,we get

(-y^3sinx+y^5cosx)dx+(3y^2cosx+5y^4sinx)dy=0
therefore

𝛗 = ∫-y^3sinx+y^5cosxdx=y^3cosx+y^5sinx+h(y)
take the derivative on 𝛗 of y,we get

𝛗_y = 3y^2cosx+5y^4sinx+h'(y)=3y^2cosx+5y^4sinx
we will have h'(y) = 0, so h(y)=C is a constant

𝛗=y^3cosx+y^5sinx+C

y^3cosx+y^5sinx+C=0
since we have

y(\frac{\pi_1}{4})=√2
solve for C

C =6

y^3cosx+y^5sinx+6=0

2
##### Term Test 1 / Re: Problem 3 (morning)
« on: October 23, 2019, 09:20:39 AM »

y''-6y'+8y=48sinh(2x)
applying the hint

y''- 6y' + 8y= 24e^{2x}-24e^{-2x}
solving the homogeneous equation

r^2-6r+8=0 ,  r_1 = 2, r_2 = 4

Yc = C_1e^2 + C_2e^4
let Yp = Axe^{2x}

Yp' = A(e^{2x} + 2xe^{2x}), Yp'' = A(2e^2x + 2(e^{2x} +2xe^{2x}))
fill in the original equation

4Ae^{2x}+4Axe^{2x} - 6A(e^{2x}+2xe^{2x})+8Axe^{2x}=24e^{2x}
solve for A

A = -12
let

Ys = Be^{-2x}

Ys' = -2Be^{-2x} , Ys'' = 4Be^{-2x}

4Be^{-2x} -6(-2Be^{-2x}) + 8Be^{-2x} = -24e^{-2x}
then we have

24B = -24, B = -1
so we have general solution

y = c1e^{2x} + c2e^{4x} -12xe^{2x}-e^{-2x}
Part(b)

y(0)=0, y'(0)=0
we get

C_1+C_2-1 =0 , 2C_1 + 4C_2 -12+2 = 0
solve for C1 and C2

C_1 = -3 , C_2 = 4

y(x) = -3e^{2x}+4e^{4x}-12xe^{2x}-e^{-2x}

3
##### Term Test 1 / Re: Problem 2 (morning)
« on: October 23, 2019, 09:14:42 AM »

xy''-(2x+1)y'+(x+1)y=0

y''-\frac{2x+1}{x}y'+\frac{x+1}{x}y=0
p(x)=(2x+1)/x

𝝻=e^{-∫-\frac{2x+1}{x}dx}
W=C𝝻

𝝻=Ce^{2x}x
W=C𝝻

W = Ce^{2x}x
let C =1,then we have

W{(y_1,y_2)}=xe^{2x}
for part b), to check that y1 is a solution then

y_1' = e^x,y_1'' = e^x

xe^x-(2x+1)e^x+(x+1)e^x=0
so y1 is indeed a solution.

W_{(y_1,y_2)}=xe^{2x}=e^xy_2'-e^xy_2

y_2'-y_2=xe^x
the integrating factor is

𝝻 = e^{-x}
multiply 𝝻 in this equation

e^{-x}y_2'-e^{-x}y_2=x

(e^{-x}y_2)'=x
take integral on both side

e^{-x}y_2=\frac{1}{2}x^2
move all the x to one side

y_2 = \frac{1}{2}x^2e^x
part c) , the general solution is

W = C_1y_1+C_2y_2

y = C_1e^x+\frac{C_2}{2}x^2e^x

y' = C_1e^x+\frac{C_2}{2}(2xe^x+x^2e^x)
we have y(1) = 0 and y(1)'=e

C_1e+\frac{C_2}{2}e=0, C_1e+\frac{C_2}{2}(2e+e)=e
solve for C1 and C2,then we have

C_1 =-\frac{1}{2},C_2 = 1

y=-\frac{1}{2}e^x+\frac{1}{2}x^2e^x

4
##### Term Test 1 / Re: Problem 4 (morning)
« on: October 23, 2019, 08:36:55 AM »

y’’- 6y'+25y=102sinx+16e^{3x}
first solving the homogenous equation

r^2-6r+25=0
solve for r, we get

r = 3 ± 4i
so we have 𝞴=3 and 𝝻=4. Then we have

y_c=c_1e^{3x}cos4x+c_2e^{3x}sin4x
let

y_p = asinx+bcosx,y_p'=acosx-bsinx,y_p'' = -asinx-bcosx
so we will have

-asinx-bcosx-6(acosx-bsinx)+25(asinx+bcosx)=102sinx
solve for a and b

a=4b,102b = 102
so a =4 , b = 1

y_{p2} = Ce^{3x},y_p2'=3Ce^{3x},y_p2'' = 9Ce^{3x}

9C-18C+25C=16
so we have C=1

y_{p2} = e^{3x}

y= c_1e^{3x}cos4x+c_2e^{3x}sin4x+e^{3x}+4sinx+cosx
next step, we are looking for the derivative of y

y'= c_1(3e^{3x}cos4x-4e^{3x}sin4x)+c_2(3e^{3x}sin4x+4e^{3x}cos4x)+3e^{3x}+4cosx-sinx
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.

c_1=-2, 3c_1+4c_2+7=0

c_1 =-2 , c_2 = -1/4

y =-2e^{3x}cos4x-\frac{1}{4}e^{3x}sin4x+e^{3x}+4sinx+cosx

5
##### Quiz-4 / quiz 4 tut 0401
« on: October 18, 2019, 02:29:22 PM »
find the general solution to the given equation

9y''+6y'+y=0
we get

9r^2+6r+1=0
solve for r, we have

3r+1=0, r = -1/3
so we have

y(t) = C1e^{-1/3}+C2te^{-1/3}

6
##### Quiz-3 / quiz 3 tut 0401
« on: October 11, 2019, 02:02:30 PM »

y'' + 3y' = 0 , y(0) = -2, y'(0) = 3
we get

r^2 + 3r = 0
Solve for r, we get the solution for r.

r1= -3, r2 = -3
For the repeated roots, the solution for y is

y(t) = C1e^{rt} + C2te^{rt}
So we have

y(t) = C1e^{-3t}+tC2e^{-3t}
Since

y(0) = -2, y'(0) = 3
We will have two equations about C1 and C2

C1 = -2, -3C1 + C2(1+0) = 3
So we have

C1 = -2, C2 = -3
Then the solution is

y(t) = -2e^{-3t}-3te^{-3t}

y(t) = - (2+3t)e^{-3t}

7
##### Quiz-2 / quiz 2 tut 0401
« on: October 04, 2019, 02:00:02 PM »

(2xy^2+2y)+(2x^2y+2x)y'=0
write y' into dy/dx, then we get

(2xy^2+2y)dx+(2x^2y+2x)dy=0
Let M = 2xy^2+2y , N = 2x^2y+2x , we get

\frac{dM}{dy}= 4xy + 2 , \frac{dN}{dx} = 4xy + 2
Since dM/dy = dN/dx, therefore they are exact. So ∃ 𝛗(x,y) = ∫ M dx

𝛗(x,y) = ∫ M dx = ∫ 2xy^2 + 2y dx = x^2y^2 + 2xy + h(y)
d𝛗(x,y)/dy = N

4x^2y + 2x +h'(y) = 2x^2y + 2x
we get h'(y) = 0 => h(y) = C, C is a constant.At the end we got the answer

𝛗(x,y) = x^2y^2 + 2xy +C

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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