Show Posts

This section allows you to view all posts made by this member. Note that you can only see posts made in areas you currently have access to.


Messages - Pengyun Li

Pages: 1 [2]
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






19
Let's see the equation first: (3x+6/y) + (x2/y +3y/x)dy/dx = 0.

Set M = 3x+6/y and N = x2/y +3y/x,
Then differentiate M w.r.t y and differentiate N w.r.t x, and then we derive that:
My= -6/(y2) and Nx = 2x/y - 3y/x2,
since My doesn't equal to Nx, then it is not exact.
Then we need to consider about the non-exact method.

However, note that:
(My-Nx)/N is not only a function of x, and (My-Nx)/M is not only a function of y.
Therefore we cannot use the normal method, so we have to use the formula:
(Nx - My)/(xM - yN) = R, where R = R(xy) depends on the quantity xy only (if you did not get it, then there must be something wrong, as if it is only function of x or only function of y, we would not need to use this method in the first place), then the differential equation had an integrating factor of the form μ(xy).

R = (Nx - My)/(xM - yN)
   = (2x/y - 3y/x2 + 6/y2) / (3x2 + 6x/y - x2 - 3y2/x)
   = 1/(xy)

Let t = xy, so R = 1/(xy) = (1/t)

μ(xy) = e(∫1/t dt) = e(lnt) = t = xy

Multiply μ(xy) on both sides of (3x+6/y) + (x2/y +3y/x)dy/dx = 0.

Then we derive that (3x2y+ 6x) + (x3 + 3y2) dy/dx = 0.

Set M = 3(x2)y+ 6x and N = x3 + 3y2,
Similarly, after differentiation, we can get that:
My = 3x2 = Nx, which is exact now!

So we can use the exact method.

There exists φ(x,y) s.t. φx = M, φy = N
Hence, φ = ∫M dx = ∫3x2y + 6x dx = yx3 + 3x2 + h(y)
(I write h(y) as it acts like a constant after we integrate w.r.t x only)
Therefore φy = x3 + h'(y) = x3 + 3y2
hence h'(y) = 3y2, h(y) = y3 + c

Therefore, φ = yx3 + 3x2 + y3 = c

Hope it helps :)






20
MAT244--Lectures & Home Assignments / Re: Problem 10 in ex 2.1
« on: October 01, 2018, 11:06:04 AM »
Hi! I have attached the detailed answer in the image. Hope it helps :)

Pages: 1 [2]