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1

Term Test 1 / Re: Problem 2 (main)

« on: October 23, 2019, 07:46:30 AM »

a)
x²y''−2xy'+(x²+2)y=0
y''-2/xy'+(1+2/x²)y=0
W=ce^(∫(2/x)dx)=cx²
Let c=1, then W=x²

b)
y1(x)=x\cos(x)
y1'=\cosx-x\sinx
y1''=-\sinx-\sinx-x\cosx
x²(-\sinx-\sinx-x\cosx)-2x(\cosx-x\sinx)+(x²+2)(x\cos(x)=0
Therefore, y1(x)=x\cos(x) is a solution of this equation.
W=|y1  y2| = |x\cos(x)                  y2| = x²
     |y1' y2'|    |\cosx-x\sinx  y2'|
Thus, x\cos(x)y2'-(\cosx-x\sinx)y2 = x²
y2'-(\cosx-x\sinx)/(x\cosx)y2 = x²/(x\cosx)
y2'-(1/x-\tanx)y2 = x²/(x\cosx)
μ= e^∫[\tanx-(1/x)]dx = e^(ln(\secx))*e^(ln(1/x)) = (\secx)/x = 1/(x\cosx)
Multiply both sides by μ
y2'/(x\cosx)-[(\cosx-x\sinx)/(x\cosx)²]y2 = [x/(\cosx)]/(x*\cosx)= 1/cos²x
[y2*1/(x\cosx)]' = 1/cos²x
y2*1/(x\cosx) = ∫[1/(\cos²x)]dx = \tanx + C
y2 = \tanx*x*\cosx = x*\sinx

c)
General solution: Y = c1y1+c2y2 = c1x*\cosx + c2x*\sinx
Y(π/2)=c2*π/2=1
c2=2/π
Y'(π/2)=-c1π/2+c2=0
c1=4/π
Thus, Y = 4/π(x*\cosx) + 2/π(x*\sinx)

2

Quiz-4 / TUT0602 Quiz 4

« on: October 18, 2019, 01:19:23 PM »

Q: 4y''+9y=0

A:
4r^2+9=0
r^2=-9/4
r=3/2*i
λ=0, μ=3/2
y(t) = c1cos(3/2*t)+c2sin(3/2*t)

3

Quiz-3 / TUT0602 Quiz 3

« on: October 11, 2019, 01:19:54 PM »

Q: Find the Wronskian of two solutions of the given differential equation without solving the equation
cos(t)y''+sin(t)y'-ty=0

Answer:
y''+sin(t)y'/cos(t)-ty/cos(t)=0
P(t)=sin(t)/cos(t)
W=ce^(∫p(t)dt)
=ce^(-∫(sin(t)/cos(t))dt)  let u=cos(t), du=-sin(t)dt
=ce^(∫1/u du)
=ce^(lnu)
=cu
=ccos(t)

4

Quiz-2 / TUT0602 Quiz2

« on: October 04, 2019, 03:47:34 PM »

Q: x²y³+x(1+y²)y′=0,  µ(x,y)=1/(xy³)
A:
Let M=x²y³, N=x(1+y²)
My=2x²y², Nx=1+y²
Since My≠Nx, this equation is not exact.
Then, multiple both sides by µ(x,y)=1/(xy³)
We have, x+(1/y³+1/y)y′=0
Let A=x, B=1/y³+1/y
Ay=Bx=0, thus, this equation now is exact.
therefore, ∃Ф(x,y), s.t. Фx=A, Фy=B
Фx=A=x
Ф(x,y)=∫xdx=1/2x²+h(y)
Фy=h′(y)=B=1/y³+1/y
h(y)=∫(1/y³+1/y)dy=-1/(2y²)+ln|y|+C
Ф(x,y)=1/2x²-1/(2y²)+ln|y|+C
General solution: 1/2x²-1/(2y²)+ln|y|=C