Differential equations – Further Pure Maths 2

Candidates should be able to:

• find an integrating factor for a first order linear differential equation, and use an integrating factor to find the general solution (e.g. \(\frac{dy}{dx} – 2y = x^2 , x \frac{dy}{dx} – y = x^4 , \frac{dy}{dx} + y \coth x = \cosh x\).)
• recall the meaning of the terms ‘complementary function’ and ‘particular integral’ in the context of linear differential equations, and recall that the general solution is the sum of the complementary function and a particular integral
• find the complementary function for a first or second order linear differential equation with constant coefficients (For second order equations, including the cases where the auxiliary equation has distinct real roots, a repeated real root or conjugate complex roots.)
• recall the form of, and find, a particular integral for a first or second order linear differential equation in the cases where a polynomial or \(a e^{bx}\) or \(a \cos px + b \sin px\) is a suitable form, and in other simple cases find the appropriate coefficient(s) given a suitable form of particular integral (e.g. evaluate \(k\) given that \(kx \cos 2x\) is a particular integral of \(\frac{d^2y}{dx^2} + 4y = \sin 2x\).)
• use a given substitution to reduce a differential equation to a first or second order linear equation with constant coefficients or to a first order equation with separable variables (e.g. the substitution x = et to reduce to linear form a differential equation with terms of the form \(ax^2 \frac{d^2y}{dx^2} + bx \frac{dy}{dx} + cy\), or the substitution \(y = ux\) to reduce \(\frac{dy}{dx} = \frac{x + y}{x – y}\) to separable form.)
• use initial conditions to find a particular solution to a differential equation, and interpret a solution in terms of a problem modelled by a differential equation.

---

Differential equations – Further Pure Maths 2 Video (5 Parts)

* For online viewing only, No Download or Print allowed *