Math Is Fun Forum
  Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °

You are not logged in.

#1 2005-11-08 07:52:42

Matilde
Member
Registered: 2005-10-30
Posts: 6

Differential equations

Can anyone see if this is right?

x * y’ - x^4 * cosx = 3y,   y(2pi)= 0

x * y’ – 3y = x^4 cosx                 | :x
y’ – (3/x) *y = x^3 * cosx

f(x) = -3/x --> F(x) =  ∫ -3/x * dx = -3ln|x| + C
the integration factor is: e^F(x) = e^-3ln|x| = 1/e^3ln|x| = 1/x^3

y’ – (3/x) *y = x^3 * cosx          | * 1/x^3
y’ * 1/x^3 – 3/x^4 * y = cosx
∫ (y’ * 1/x^3)’ =  ∫ cosx
y * 1/x^3 = sinx + C                  | :1/x^3)
y(x) = x^3 * sinx + C * x^3
y(2pi) = (2pi)^3 * sin*2pi + C * (2pi)^3 = 0

y(x)= x^3 * sinx

Can somone please help me with this one:
(x^2 + 4)y’ + 3xy = x      ; y(0)=1

I started like this
y’(3x/x^2+4)*y= x/x^2+4
f(x)=3x/x^2+4 --> F(x)  ∫ (3x/x^2+4 )dx=    and now?
Integration factor is? hmm

Matilde

Offline

#2 2005-11-08 08:18:22

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

Re: Differential equations

The one that you've done is perfect. That or we're both wrong.

With the second one, use the identity that ∫ (f'(x)/f(x))dx = ln f(x)

Yours is 3x/(x² + 4), so to get the numerator to be the differential of the denominator, you need to change it into 1.5(2x) and make the 1.5 a constant. 1.5∫ 2x/(x² + 4)dx = 1.5 ln (x² + 4), or ln (x² + 4)^1.5

As the integrating factor is e^f(x), the e and ln cancel out to leave you with (x² + 4)^1.5. I'll leave you to try the rest of it yourself, feel free to come back if you get stuck again.


Why did the vector cross the road?
It wanted to be normal.

Offline

#3 2005-11-08 23:53:11

Matilde
Member
Registered: 2005-10-30
Posts: 6

Re: Differential equations

y´ + (3x/(x^2 + 4))y = x/(x^2 + 4)

Int. Factor is: e^[int_(3x/(x^2 + 4))] 
= e^[(3/2)*ln(x^2 + 4)] = e^[ln((x^2 + 4)^(3/2))] = (x^2 + 4)^(3/2).


(x^2 + 4)^(3/2)y´+ 3x(x^2 + 4)^(1/2) = x(x^2 + 4)^(1/2)

[(x^2 + 4)^(3/2)y]´ = x(x^2 + 4)^(1/2)

(x^2 + 4)^(3/2)y
= int_[x(x^2 + 4)^(1/2) dx] (u=x^2 + 4)
= int_[u^(1/2)/2]
= (u^(3/2)/3) + C 
= ((x^2 + 4)^(3/2)/3) + C (3)


y = (1/3) + C(x^2 + 4)^(-3/2).

y(0)=1 --> = (1/3) + C*(4)^(-3/2) = (1/3) + (C/8),  C=8*(1 - (1/3)) = 8*2/3 = 16/3. 

y = [16(x^2 + 4)^(-3/2) + 1]/3.

Thanx mathsyperson! smile

Offline

Board footer

Powered by FluxBB