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### Messages - Pengyun Li

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16
##### Thanksgiving Bonus / Re: Thanksgiving bonus 1
« on: October 05, 2018, 08:52:24 PM »
We want to find a second order equation with the fundamental system of solutions $\{y_1(x),y_2(x)\} = \{\frac{1}{x+1}, \frac{1}{x-1}\}$.

Then $y_1=\frac{1}{x+1}, y_2=\frac{1}{x-1}$.

$\implies$ $W(y,y_1,y_2) = W(y, \frac{1}{x+1},\frac{1}{x-1})$ = $\left|\begin{matrix}y & \frac{1}{x+1} & \frac{1}{x-1}\\ y' &-\frac{1}{(x+1)^2} &-\frac{1}{(x-1)^2}\\ y'' &\frac{2x}{(x+1)^4}& \frac{2x}{(x-1)^4} \end{matrix}\right|= 0.$

$\implies$ Solve the determinant : $y\ \left|\begin{matrix} -\frac{1}{(x+1)^2} & -\frac{1}{(x-1)^2}\\ \frac{2x}{(x+1)^4} &\frac{2x}{(x-1)^4}\end{matrix}\right| - y^{'}\ \left|\begin{matrix} \frac{1}{x+1} & \frac{1}{x-1}\\ \frac{2x}{(x+1)^4} &\frac{2x}{(x-1)^4}\end{matrix}\right| + y^{''}\ \left|\begin{matrix} \frac{1}{x+1} & \frac{1}{x-1}\\ -\frac{1}{(x+1)^2} &-\frac{1}{(x-1)^2}\end{matrix}\right|$

= $y\ (-\frac{8x^2}{(x+1)^4(x-1)^4}) - y^{'}\frac{12x^3+4x}{(x+1)^4(x-1)^4} + y^{''}\frac{-2}{(x+1)^2(x-1)^2} = 0$

Wrong second derivatives!

Then we multiply both sides with $(x-1)^4(x+1)^4$ to get:

$y(-8x^2) - y^{'}(12x^3+4x) + y^{''}(-2x^4+4x^2-2) = 0$

our final equation is: $(-x^4+2x^2-1)y^{''}-(6x^3+2x)y^{'}+(-4x^2)y = 0$.

17
##### Quiz-2 / Re: Q2 TUT 0401 and TUT 0601
« on: October 05, 2018, 06:14:52 PM »
(x2y3)+x(1+y2)y'= 0

Let M = x2y3, N = x(1+y2)

My = $\frac{d(M)}{dy}$ = 3x2y2

Nx = $\frac{d(N)}{dx}$ = 1+y2

Since My ≠ Nx, hence not exact, and thus we need to use the integrating factor μ = 1/(xy3)

Multiply μ on both sides of the original equation,

1/(xy3)x2y3 + 1/(xy3)x(1+y2)y' = 0

Now let M' = 1/(xy3)x2y3 = x,

N' = 1/(xy3)x(1+y2) = y-3 + y-1,

M'y = $\frac{d(M')}{dy}$ = 0,

N'x = $\frac{d(N')}{dx}$ = 0,

M'y = N'x, hence exact now.

There exist φ(x,y) s.t. φx = M', φy = N'

φ(x,y) = ∫M'dx =$\frac{1}{2}$x2 + h(y)

φy = h'(y) = N' = y-3 + y-1

Thus h'(y) = y-3 + y-1, h(y) = -$\frac{1}{2}$y-2 + ln|y| + C

Therefore, φ(x,y) =$\frac{1}{2}$x2 -$\frac{1}{2}$y-2 + ln|y|= C

18
##### Quiz-2 / Re: Q2 TUT 0201
« on: October 05, 2018, 05:54:21 PM »
(x+2)sin(y) + xcos(y)y' = 0

Let M = (x+2)sin(y), N = xcos(y)

My = $\frac{d((x+2)sin(y))}{dy}$ = (x+2)cos(y)

Nx = $\frac{d(xcos(y))}{dx}$ = cos(y)

Since My ≠ Nx, hence not exact, and thus  we need to find the integrating factor.

Let $\frac{My - Nx}{N}$. we can derive $\frac{(x+2)cos(y)-cos(y)}{cos(y)}$ = $\frac{x+1}{x}$, which is a function of x only.

μ = e∫$\frac{x+1}{x}$dx = xex, which is the integrating factor.

Multiply μ on both sides of the original equation,

xex(x+2)sin(y) + x2cos(y)exy' = 0

Now let M' = xex(x+2)sin(y), N' = x2cos(y)ex,

M'y = $\frac{d(M')}{dy}$ = (x+2)xexcos(y),

N'x = $\frac{d(N')}{dx}$ = (x+2)xexcos(y),

M'y = N'x, hence exact now.

There exist φ(x,y) s.t. φx = M', φy = N'

φ(x,y) = ∫M'dx =x2exsin(y) + h(y)

φy = φ'(x,y) = x2excos(y)+h'(y) = N'

x2excos(y) + h'(y) = x2excos(y)

Thus h'(y)=0, h(y) is a constant.

Therefore, φ(x,y) = x2exsin(y) = C

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