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## #1 2005-11-09 06:52:42

Matilde
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### 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

(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

## #2 2005-11-09 07:18:22

mathsyperson
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### 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.

## #3 2005-11-09 22:53:11

Matilde
Novice

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### 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!