Gday, Just trying to work out the following linear differential equation. I'm trying to work it out via the Integrating Factor method and I am getting stuck at the point where you multiply both sides by I(x). The equation is: xy' - 2y = x^2 Can anyone help me out with this? Regards, Matt PS. Also, how do I type equations, etc in the math font?
Given: \(xy' - 2y = x^2\) Divide by x to make it look like an integrating factor problem: \(y' - (2/x) y = x \) Multiply by the (as of yet unknown integrating factor): \(f(x)y' + f(x)(-2/x) y = f(x)x\) Goal for the integrating factor to make the left side look like a simple chain product rule derivative: \((f(x) y)' = f(x)y' + f'(x)y = f(x)y' + f(x)(-2/x) y \) But this means: \( f'(x) = (-2/x) f(x)\) or \( f'(x)/f(x) = -2/x \) with solution \(f(x) = \frac{A}{x^2}\) Plugging that back in to our problem we get: \(Ay'/x^2 + (A/x^2)(-2/x) y = (A/x^2)x\) which we ease wear and tear on the brain by writing as \((Ay/x^2)' = u' = A/x\) With solution \(u = Ay/x^2 = A \ln x + B\) or \(y = x^2 \ln x + \frac{B}{A} x^2 = x^2 \ln x + C x^2 \) Solution: \(y = x^2 \ln x + C x^2\) Check: \(y' = x^2/x + 2 x \ln x + 2 C x \) \(xy' - 2y = x^2 + 2 x^2 \ln x + 2 C x^2 - 2 x^2 \ln x - 2 C x^2= x^2\)
Ah okay, thanks for your reply. One thing I am not understanding is the multiplication into the equation of the integrating factor. It has confused me on equations other than this one also. Here is another example of an equation that is confusing me: 1/x . dy/dx - y/x^2 = 1 becomes d/dx (y/x) = 1 How does the breaking down of the left side work? Where did y/x come from? Matt
Multiplying through by an integrating factor makes the left hand side an exact derivative. To use your questions as an example \(x \frac{dy}{dx} - y\) is not an exact derivative. However, a multiple of it is but you might not be able to spot it. Hence you use the integrating factor, which is the multiple you're looking for. If you multiply your expression by \(\frac{1}{x^{2}}\) you do get an exact derivative, \(x \frac{dy}{dx} - y\) goes to \(\frac{1}{x} \frac{dy}{dx} - \frac{y}{x^{2}}\). This is an exact derivative \(\frac{d}{dx}\left( \frac{y}{x} \right)\) and so you can integrate it easily. That's the utility of such a method.
[thread=61223]Displaying equations using Tex[/thread] Example: \(\frac{-b\pm\sqrt{b^2-4ac}}{2a}\)Displays as: \(\frac{-b\pm\sqrt{b^2-4ac}}{2a}\) You can also click the "quote" button on any post containing an equation to see how that equation was made.
