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Matilde
2005-11-09 22:53:11

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!

mathsyperson
2005-11-09 07:18:22

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.

Matilde
2005-11-09 06:52:42

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