Edexcel January 2007 - Question 4 (Rates and Differential Equations)

Rates and differential equations is a topic many students find difficult, especially when solving the differential equation is combined with techniques from other topics on the C4 syllabus. I hope my walkthrough of this particular exam question is helpful:

4(a). Express (2x-1)/(x-1)(2x-3) in partial fractions.

Our fraction splits into distinct linear factors and so we are looking to find constants A,B such that

(2x-1)/(x-1)(2x-3) = A/(x-1) + B/(2x-3)

There are many shortcuts to calculate partial fraction constants, however I prefer to explicitly write out the steps in order to ensure I would get the method marks in the case of making a mistake in calculation. We manipulate the right hand side to be a single fraction and then equate the numerators of either side to attain:

2x-1= A(2x-3) + B(x-1)

Then equating the x coefficient and constant terms yields two equations in A and B to be solved simultaneously:

2A+B=2 and -3A-B=-1 =====> A=-1 and B=4

4(b). Given that x>=2, find the general solution of the differential equation (2x-3)(x-1)dy/dx= (2x-1)y

Here it is useful to notice the similarity between the factors in the differential equation and the fraction in part (a). This differential equation also can be seen to be separable, and x>=2 means we do not need to be concerned with dividing by (x-1)(2x-3) as this is never zero:

1/y dy=(2x-1)/(x-1)(2x-3) dx

We now use the partial fractions we calculated in part (a) since we know how to integrate terms of the form a/(bx-c).

1/y dy= -1/(x-1) + 4/(2x-3) dx

Integrating both sides we attain:

log(y) = -log(x-1) + 2log(2x-3) + log(C)

Where we are careful to remember the constant log(C), and that the coefficient of x in the second fraction needs to be considered. Next we simplify the right hand side using standard log properties:

log(y) = log(C(2x-3)^2/(x-1))

Thus applying exp to both sides we attain:

y = C(2x-3)^2/(x-1) as our general solution.

4(c). Hence find the particular solution of this differential equation that satisfies y = 10 at x = 2, giving your answer in the form y = f(x).

The final part of this question is a matter of finding the particular constant C for the (x,y) value they specify. Substituting their values into our general solution we get:

10 = C(2*2-3)^2/(2-1) = C

And so C = 10 in the particular solution. Fortunately our solution is already in the desired form y = f(x):

y = 10(2x-3)^2/(x-1)