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FP3 - Second Order DE - Sub Prob - AQA

rosschambers1987

Do I treat the ux as two variable. I'm not sure if it means u to be a constant, although I doubt it because they would have otherwise used "a".

If I want to differentiate ux with respect to x do I use the product rule some how?

Like would it be dy/dx = du/dx + 1?

So then d^y/dx^2 = d^u/dx^2?

Cheers,
Ross

Reply 1

Dr Grip

The point of that substitution is to make the equation easier to solve. So u and x are treated as two variables. And yes, you do use product rule to differentiate y=ux, but the answers should be

\frac{dy}{dx}=x\frac{du}{dx}+u

\frac{d^2y}{dx^2}=\frac{du}{dx}+x\frac{d^2u}{dx^2}+\frac{du}{dx}

Not sure if you just did a silly mistake there or not. :biggrin:

Reply 2

mr tim

remember in product rule, you leave one bit fixed and differentiate the other, and then do it the other way round. Both things are then added together.

Yes ux are both variables. If you aren't sure, you see the du/dx in the equation, suggesting that u is a variable.