dif equations

**moolimanj** · February 14th 2008, 12:58 AM

Hi all

which method do i use to solve this:

**mr fantastic** · February 14th 2008, 01:24 AM

Originally Posted by moolimanj

Hi all

which method do i use to solve this:

Integrating factor method.

But I'd first re-arrange into: $\frac{dy}{dx} + \frac{2}{x} \, y = \frac{5}{x^3}$ .

Hint: The integrating factor is $e^{\int \frac{2}{x} \, dx} = e^{2 \ln x} = e^{\ln x^2} = x^2$ .

**dankelly07** · February 15th 2008, 08:13 AM

sorry to butt in..but I want to understand differential equations better, if you use the integrating factor do you multiply by all the terms? how would you proceed from here?

**topsquark** · February 15th 2008, 09:42 AM

Originally Posted by dankelly07

sorry to butt in..but I want to understand differential equations better, if you use the integrating factor do you multiply by all the terms? how would you proceed from here?

Multiply both sides by $x^2$ . That's the only way to make sure the equation still says the same thing.
$\frac{dy}{dx} + \frac{2}{x} \, y = \frac{5}{x^3}$

$x^2 \cdot \frac{dy}{dx} + x^2 \cdot \frac{2}{x} \, y = x^2 \cdot \frac{5}{x^3}$

$x^2 \cdot \frac{dy}{dx} + 2xy = \frac{5}{x}$

$\frac{d}{dx} \left ( x^2y \right ) = \frac{5}{x}$

Now integrate both sides with respect to x:
$x^2y = 5~ln(x) + C$

You take it from here.

-Dan

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