Solving Integrating Factor Homework Statement

[gtfitzpatrick]

Homework Statement

use an integrating factor to solve
<br /> \frac{ \partial u}{ \partial x} = -2 + \frac{u}{2x} <br />

The Attempt at a Solution

let P(x) = \frac {1}{2x}

M(x) = e^(\int(\frac {1}{2x}dx))

= \sqrt{x}

so u = <br /> \frac{1}{ \sqrt{x}} \int-2\sqrt{x} dx = \frac{1}{ \sqrt{x}}(\frac{2}{3} x^(3/2))<br />

=-4x + k


anyone got any ideas if I am doing this right?
thanks

 

[fzero]

You have a numerical integration factor in the wrong place

so u = <br /> \frac{1}{ \sqrt{x}} \int-2\sqrt{x} dx = \frac{1}{ \sqrt{x}}(\frac{2}{3} x^{(3/2)} + \mathbf{k})<br />

=-4x + \mathbf{\frac{k}{\sqrt{x}}}

 

[gtfitzpatrick]

deadly thanks a mill

 

[gtfitzpatrick]

sorry should it be <br /> =\frac {-4x}{3} + \mathbf{\frac{k}{\sqrt{x}}}<br />

 

[fzero]

Sorry, I looked at the -4x in your solution, which is correct, so I missed an earlier mistake. You should have had

P(x) = -\frac{1}{2x}.

This changes quite a bit, so you should go back through the algebra.

 

[gtfitzpatrick]

i don't understand why <br /> P(x) = -\frac{1}{2x}.<br /> should it not be term infront of u?

 

[fzero]

i don't understand why <br /> P(x) = -\frac{1}{2x}.<br /> should it not be term infront of u?

P(x) is defined with that term on the same side as the derivative:

u&#039;(x) + P(x) u(x) = q(x).

Remember that the integrating factor works because

(u&#039;(x) + P(x) u(x) ) \exp \left(\int P(x) dx\right) = \frac{d}{dx} \left[ u(x) \exp \left(\int P(x) dx \right) \right].

 

[gtfitzpatrick]

ahh yes i didnt bring it across, thanks

 

[gtfitzpatrick]

i think this is right?
let P(x) = \frac {-1}{2x}

M(x) = e^(\int(\frac {-1}{2x}dx))

= \frac{1}{\sqrt{x}}

so u = <br /> \sqrt{x} \int \frac {-2}{\sqrt{x}} dx = \sqrt{x}(-4\sqrt{x} + k)<br />

=-4x + \sqrt{x} k

 

[fzero]

The easiest way to verify a solution to a differential equation is to explicitly check that it actually satisfies the differential equation.

 

[gtfitzpatrick]

i can't figure out where i went wrong!

 

[dextercioby]

You didn't go wrong. I found the same solution

u(x)= -4x + C\sqrt{x}

 

[gtfitzpatrick]

oh, ok cool. thanks a mill for your help...you have great patience