Messages

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)

M_y = $\frac{d(M)}{dy}$ = 3x²y²

N_x = $\frac{d(N)}{dx}$ = 1+y²

Since M_y ≠ N_x, 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. φ_x = M', φ_y = N'

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

φ_y = 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)

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

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

Since M_y ≠ N_x, 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} = xe^x, which is the integrating factor.

Multiply μ on both sides of the original equation,

xe^x(x+2)sin(y) + x²cos(y)e^xy' = 0

Now let M' = xe^x(x+2)sin(y), N' = x²cos(y)e^x,

M'_y = $\frac{d(M')}{dy}$ = (x+2)xe^xcos(y),

N'_x = $\frac{d(N')}{dx}$ = (x+2)xe^xcos(y),

M'_y = N'_x, hence exact now.

There exist φ(x,y) s.t. φ_x = M', φ_y = N'

φ(x,y) = ∫M'dx =x²e^xsin(y) + h(y)

φ_y = φ'(x,y) = x²e^xcos(y)+h'(y) = N'

x²e^xcos(y) + h'(y) = x²e^xcos(y)

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

Therefore, φ(x,y) = x²e^xsin(y) = C