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ConfusedMathsStudent

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Differentiating Differential Equations

Hello, I have a question which asks me to differentiate a differential equation. I have no idea how you do this.

I have xdy/dx + y = 1/x²

How would you differentiate that?

Could you rearrange it to get:

dy/dx = (1/x²-y)/x = (1-yx²)/x³

How would you differentiate that?

Thank you for your help!

It is greatly appreciated.

An

Guest

Re: Differentiating Differential Equations

There is a general result you can use.

f(x)dy/dx + f'(x)y = d(f(x)y)/dy

So for your equation,
f(x) = x    then f'(x) = 1

Sub that into the genral equation, so

xdy/dx +y = d(xy)/dx

d(xy)/dx = 1/x^2

xy = the integral of 1/x^2 with respect to x

then if you integrate 1/x^2 you get -1/(2x^3) + c, so

xy = -1/(2x^3)

y = -1/(2x^4) + c/x

I hope that makes sense. If not, just search for "first order differential equations"

Bob

Administrator

Registered: 2010-06-20

Posts: 10,649

Re: Differentiating Differential Equations

hi ConfusedMathsStudent

(note to An : you seem to be integrating the DE rather than differentiating
Also, when you integrate 1/x², raise the power to -1 and divide by the new power ie -1/x  ?)

I assume the question requires that you differentiate again with respect to x.

Your re-arrangement is a perfectly good way to begin.

So you have a 'quotient' with the top item (=u) containing the product yx².

If you've not had that before it goes like this:

Put that together with the quotient rule and you get to a result.  It'll have dy/dx in it.

Post again if you need more details than this.

Bob

Last edited by Bob (2011-03-31 19:02:33)

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