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## #1 Re: Help Me ! » differential equations » 2005-11-14 01:07:48

Hi Mathsyperson!
I know that the answer is e^((-9/2)x)  (Acos(3√7/2)x) + Bsin(3√7/2)x)

## #2 Help Me ! » differential equations » 2005-11-11 02:09:26

Matilde
Replies: 3

How can I solve this:

y + 9y' + 36y= 0

λ^2 + 9λ + 36= 0 → λ= ?

Matilde

## #3 Re: Help Me ! » Differential equations » 2005-11-08 23:53:11

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!

## #4 Help Me ! » Differential equations » 2005-11-08 07:52:42

Matilde
Replies: 2

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

(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?

Matilde

## #5 Re: Help Me ! » Mathhelp! » 2005-11-06 22:58:59

I know Im late.. but ThankYouVeryMuch!

## #6 Re: Help Me ! » Mathhelp! » 2005-10-30 07:01:29

Thanx kylekatarn!

I have to solve:

Volume=   ∫ π (e^2x * (sinx)^2 * dx ,0 ,π

I know that the integral of e^2x = 1/2 e^2x.... but I don't know what to do with (sinx)^2?