differential equations

**twilightstr** · March 1st 2009, 03:57 PM

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)

**Prove It** · March 1st 2009, 04:05 PM

Originally Posted by twilightstr

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)

Use the integrating factor method.

Divide everything by $x$ and the equation becomes

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

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

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

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

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

Can you go from here?

**twilightstr** · March 1st 2009, 04:32 PM

[quote=Prove It;274960]Use the integrating factor method.

Divide everything by $x$ and the equation becomes

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

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

how did u get the above step

**Prove It** · March 1st 2009, 08:55 PM

[quote=twilightstr;274978]

Originally Posted by Prove It

Use the integrating factor method.

Divide everything by $x$ and the equation becomes

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

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

how did u get the above step

It's a standard technique for first order linear DE's.

If you have a first order linear DE of the form

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

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

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

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

which is usually solvable.

**HallsofIvy** · March 2nd 2009, 04:45 AM

Originally Posted by twilightstr

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)

It will help you considerably to recognise that $e^{2ln x}= e^{ln x^2}= x^2$ !
y= $\frac{\int^x t^4 dt}{x^2}$

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.)

**Prove It** · March 2nd 2009, 12:36 PM

Originally Posted by HallsofIvy

It will help you considerably to recognise that $e^{2ln x}= e^{ln x^2}= x^2$ !
y= $\frac{\int^x t^4 dt}{x^2}$

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.)

You have to divide everything through by x first.

So $P(x) = -\frac{2}{x}$ , not -2.

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