solve the differential equations

xy'-2y = x^2

my work;

dy/dx = x+ 2y/x

p(x)= 2/x

Q(x)= x^2

integral 2/x = 2lnx

Ae^2lnx

y= (integral e^2lnx)(x^2)/(e^2lnx)

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- Mar 1st 2009, 03:57 PMtwilightstrdifferential equations
solve the differential equations

xy'-2y = x^2

my work;

dy/dx = x+ 2y/x

p(x)= 2/x

Q(x)= x^2

integral 2/x = 2lnx

Ae^2lnx

y= (integral e^2lnx)(x^2)/(e^2lnx) - Mar 1st 2009, 04:05 PMProve It
Use the integrating factor method.

Divide everything by $\displaystyle x$ and the equation becomes

$\displaystyle \frac{dy}{dx} -\frac{2}{x}y = x$.

The integrating factor is $\displaystyle e^{\int{-\frac{2}{x}\,dx}} = e^{-2\ln{x}} = e^{\ln{(x^{-2})}} = x^{-2}$.

So multiply everything by $\displaystyle x^{-2}$ and you get

$\displaystyle x^{-2}\frac{dy}{dx} - 2x^{-3}y = x^{-1}$

$\displaystyle \frac{d}{dx}(x^{-2}y) = x^{-1}$.

Can you go from here? - Mar 1st 2009, 04:32 PMtwilightstr
[quote=Prove It;274960]Use the integrating factor method.

Divide everything by $\displaystyle x$ and the equation becomes

$\displaystyle \frac{dy}{dx} -\frac{2}{x}y = x$.

The integrating factor is $\displaystyle e^{\int{-\frac{2}{x}\,dx}} = e^{-2\ln{x}} = e^{\ln{(x^{-2})}} = x^{-2}$.

how did u get the above step - Mar 1st 2009, 08:55 PMProve It
[quote=twilightstr;274978]It's a standard technique for first order linear DE's.

If you have a first order linear DE of the form

$\displaystyle \frac{dy}{dx} + P(x)y = Q(x)$

The integrating factor is $\displaystyle I(x) = e^{\int{P(x)\,dx}}$.

Multiply both sides of the DE by the integrating factor and it reduces to

$\displaystyle \frac{d}{dx}(I(x)y) = I(x)Q(x)$

which is usually solvable. - Mar 2nd 2009, 04:45 AMHallsofIvy
It will help you considerably to recognise that $\displaystyle e^{2ln x}= e^{ln x^2}= x^2$!

y= $\displaystyle \frac{\int^x t^4 dt}{x^2}$

That is easy to integrate. (I wrote "t" for the dummy variable inside the integral rather than "x" to remind you that you cannot cancel the "x" in the denominator until after you have integrated.) - Mar 2nd 2009, 12:36 PMProve It