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First Order Differential Equation

How can you simplify the following into d/dx[(some function here)]?

\frac{1}{x}\frac{dy}{dx} - \frac{2}{x^2}y

?

I mean I know it's something along the lines of

\frac{d}{dx}(\frac{y}{x})

but the 2 in front of the y is really excruciatingly difficult to deal with!

I'm blatantly just going to cry when someone shows me what I'm doing wrong.

Reply 1

+ polarity -

It looks like an Integrating Factor question to me... but I can't help you. :o:

Reply 2

wanderlust.xx

Well the original question was in fact

x\frac{dy}{dx} = 2y + (x^4)(e^(-x))

So I took the 2y to the other side and found my integrating factor, which was

\frac{1}{x^2}

. Multiply the original equation by the integrating factor and you get what I posted in the original post.

Reply 3

Vikki =]

then just do d/dx(Iy) = IQ
and integrate both sides

That's what I'd do anyway...

Reply 4

nuodai

wanderlust.xx

Well the original question was in fact

x\frac{dy}{dx} = 2y + (x^4)(e^(-x))

So I took the 2y to the other side and found my integrating factor, which was

\frac{1}{x^2}

. Multiply the original equation by the integrating factor and you get what I posted in the original post.

It would help to divide through by

x

, then you end up with

\dfrac{1}{x^2}\dfrac{dy}{dx} - \dfrac{2}{x^3} y

, which you can do something with.

When you're doing questions like this, make sure that dy/dx has nothing multiplying it -- if it does, divide through by it.

For what it's worth, if

\mu(x)

is your integrating factor (i.e.

\mu(x) = \dfrac{1}{x^2}

in this case), then you should end up with

\dfrac{d}{dx} \big( \mu(x)y \big)

on the LHS. This only works when there is nothing multiplying dy/dx though, so be careful!

Reply 5

Pheylan

\frac{dy}{dx}-\frac{2}{x}y=\frac{x^3}{e^x}

e^{-2\int\frac{1}{x}\ dx}=e^{-2lnx}=x^{-2}

x^{-2}\frac{dy}{dx}-\frac{2}{x^3}y=xe^{-x}

wanderlust.xx

Well the original question was in fact

x\frac{dy}{dx} = 2y + (x^4)(e^(-x))

So I took the 2y to the other side and found my integrating factor, which was

\frac{1}{x^2}

. Multiply the original equation by the integrating factor and you get what I posted in the original post.

you should've divided by x before finding the IF

Reply 6

wanderlust.xx

I did divide through by x before finding the integrating factor. Don't I multiply my original equation by the integrating factor?

Wait... I'm an idiot.