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)

Answered by Oliver V. Maths tutor

10403 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

a) Point A(6,7,2) lies on l1. Point B(9,16,5) also lies on l1. Find the distance between these two points. b) l2 lies in the same z plane as l1 and crosses l1 at A and is perpendicular to l1. Express l2 in vector form.


Consider the unit hyperbola, whose equation is given by x^2 - y^2 = 1. We denote the origin, (0, 0) by O. Choose any point P on the curve, and label its reflection in the x axis P'. Show that the line OP and the tangent line to P' meet at a right angle.


How would I find the approximate area enclosed by the expression e^x*sin(x)*x^3 on an infinite scale?


Integrate cos^2(x)


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences