Question

Solve the first order IVP.
dy/dx +P(x)y=x^-6; y(1)=-1

P(x)={5/x, 1<_x<_e}
P(x)={-1/x, x>e}

Answer 1

For 1st condition: 1<=x<e
e^int(P(x)) = x^5
x^5(dy/dx) + 5x^4y = x^-1
(x^5y)' = x^-1
integrate both sides
x^5y = ln(x) +C
y = ln(x)/x^5 + C
y(1) = -1
-1 = ln(1)/1 + C
-1 = C

y = (ln(x)/x^5) -1

Answer 2

not sure if you need to work out 2nd condition since initial value has already been satisfied

Answer 3

How did you get ***=ln(x)+c?
Shouldn't it be -1/5 x^-5 +c?

Answer 4

oh i multiplied the x^-6 by x^5 resulting in x^-1

Answer 5

im pretty sure you multiply the x^5 on both sides to keep equation balanced

Answer 6

oh i multiplied the x^-6 by x^5 resulting in x^-1

Answer 7

Dang it, I always forget that.
Is that the correct form for the equation for both of them, because I remember my professor saying that you have to do something differently with a piecewise equation. I can't remember what though.

Answer 8

yeah thats what im not sure on either...sorry

Answer 9

Thanks anyway. Maybe I'kk post it again later...

Answer 10

Anyone else know?

Answer 11

IIRC you take the iv for the second piece by by plugging in the end of domain of first piece which is start of second (x=e?) The second piece is a different equation.

Your result will be piecewise as well...ln etc, 1<_x<_e and whatever, x>e.

Answer 12

I'm really not sure I understand what you're saying. Can you show me with the problem?

Answer 13

You got y = (ln(x)/x^5) -1 for first piece, 1<_x<_e.

Plug in e (end of domain for first piece, start of domain for second piece) to get
ln(e)/e^5 -1 = e^-5-1 and this is your "y1" for the second leg where (P(x) = -1/x).