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We'll first compute these intersections by setting x=0 and y=0 consecutively. This gives y=a-1 and a/(x-1)^2-1=0. Hence we find (x-1)^2=a, so x=1+-sqrt(a). As we have a>1 and we want the intersection w...

We know that non-real roots appear in complex conjugate pairs. Hence when we know one root, we know both of them. Then, as we can factorise a quadratic in it's linear factors, we know our quadratic is a c...

The answer is Ln8/9, by first converting (1-x)/(5x-6-x^2) into partial fractions you get 1/(2-x) + 2/(x-3), the next step is a simple integration by inspection followed by log manipulations to get the fin...

To make x the subject we want to try to get all the x's on one side of the equals sign! Firstly to get rid of the fraction we multiply both sides by 2, giving 3x+y= 2(x+z). We then expand the brackets on ...

The integral of any equation let the example be dy/dx = ax^n The integral of (RHS) dy/dx (because when we integrate we are integrating both sides) is y The integral of (LHS) ax^n is  [ax^(n+1)]/[n+1] when...