I'm struggling with an FP2 First-Order Differential Equations Question (Edexcel June 2009 Q3) and the topic in general!

(In terms of formatting, we'll use y' to represent dy/dx, exp() to represent the expoential e^(), int() to represent integration with respect to x, and 1ODE to mean First-Order Ordinary Differential Equation). 1ODE to consider: "(sin(x))y' - (cos(x))y = sin(2x)sin(x)". There are typically two methods of solving 1ODE's - seperation of variables, and using a Multiplier M(x) known as an integrating factor. The latter of these methods is what we'll use here. Firstly, let's recap. The standard form of a 1ODE is "y' + P(x)y = Q(x)", where P and Q are simply functions of x. The integrating factor is defined as M(x) = exp(int(P(x))). In order to use the integrating factor, we must have our 1ODE in the standard form (i.e. y' needs to be on it's own!). The key to solving 1ODE's is to recognise the LHS (left-hand side), after multiplication by a factor (i.e. M(x)), as the derivative of a product. We'll see how this works in the example above. We need our 1ODE in the standard form, so we'll start by dividing through by sin(x) to get y' on it's own. This gives "y' - cot(x)y = sin(2x)", remembering that cot(x)=1/(tan(x))=(cos(x))/(sin(x)). Now we have a form where we can let P(x)=-cot(x) and Q(x)=sin(2x), using the labels from our standard form equation above. Next, we need to calculate our integrating factor. Using our definition and our P(x), we get M(x)=exp(int(-cot(x)))=exp(-ln|sin(x)|)=1/(sin(x)). [Can you see how we get this? Hint: write cot(x) in terms of sin and cos, then use Integration by Substitution with u=sin(x) to get the result]. We can ignore the constant of integration here. Next, we can multiply our standard form 1ODE by the integrating factor to give "(1/(sin(x)))y' - [(cos(x))/(sin(x)sin(x))]y = 2cos(x)". The RHS has become simply "2cos(x)" because sin(2x)=2sin(x)cos(x), and the sin(x) from sin(2x) will cancel with the sin(x) in the integrating factor to leave cos(x). [Can you see this? If not, write it out and simplify!]. I said earlier that the method of solving a 1ODE is to recognise the LHS as the derivative of a product. So, let's write our 1ODE in this form. We get "d(y/(sin(x)))/dx = 2cos(x)". This is because when we evaluate the LHS (i.e. differentiate y/(sin(x)) with respect to x), we get exactly the LHS of our earlier 1ODE. [Can you see this? Try differentiating y/(sin(x)) with respect to x using the product rule]. The next step is to integrate both sides with respect to x. The LHS becomes "int(d(y/(sin(x)))/dx)", which simplifies to "y/(sin(x))" because our integral and differential both cancel. [Why? Because integration is the opposite of differentiating]. The RHS is a simple integration scenario, resulting in "2sin(x) + c". Don't forget the constant of integration (+c). So, we now have our 1ODE in the form "y/(sin(x)) = 2sin(x) + c". We can multiply both sides by sin(x) to get it in the form y=..., so we have: "y = 2(sin(x))^(2) + csin(x)". This is the general solution to our 1ODE. It is easy to see why this is an 8 mark question. There are lots of places where carelessness can create mistakes and cost marks, such as: - Not dividing the RHS by sin(x) in the first step. - Miscalculating the integrating factor, particularly by forgetting the exp() or integrating cot(x) wrong. - Forgetting the constant of integration in the general solution. (You can see how costly this is as we now have csin(x)!)

Related Further Mathematics A Level answers

All answers ▸

Prove by induction that n! > n^2 for all n greater than or equal to 4.


differentiate arsinh(cosx))


Find the equation of the tangent to the curve y = exp(x) at the point ( a, exp(a) ). Deduce the equation of the tangent to the curve which passes through the point (0,1) .


Find the values of x where x+3>2/(x-4), what about x+3>2/mod(x-4)?


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 2013–2024

Terms & Conditions|Privacy Policy