Second Order Differential Equations

Second order differential equations involve the second derivative d^2y/dx^2 and are a major topic in AQA Further Mathematics. This lesson covers homogeneous and non-homogeneous linear equations with constant coefficients, the auxiliary equation method, and finding particular integrals.

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1 General Form

A second order linear ODE with constant coefficients has the form:

a (d^2y/dx^2) + b (dy/dx) + c y = f(x)

where a, b, c are constants and f(x) is a given function.

• If f(x) = 0, the equation is homogeneous.
• If f(x) is not zero, the equation is non-homogeneous (or inhomogeneous).

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2 The Auxiliary Equation (Homogeneous Case)

For a y'' + b y' + c y = 0, try the solution y = e^(mx). Substituting:

a m^2 e^(mx) + b m e^(mx) + c e^(mx) = 0

Dividing by e^(mx) (which is never zero):

a m^2 + b m + c = 0

This is the auxiliary equation (AE). The nature of its roots determines the form of the general solution.

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Case 1: Two Distinct Real Roots (m_1 and m_2)

y = A e^(m_1 x) + B e^(m_2 x)

where A and B are arbitrary constants.

Case 2: Repeated Real Root (m_1 = m_2 = m)

y = (A + Bx) e^(mx)

The factor of x is needed because e^(mx) alone gives only one independent solution.

Case 3: Complex Conjugate Roots (m = p + qi and p - qi)

y = e^(px) (A cos qx + B sin qx)

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3 Worked Example 1 — Two Distinct Real Roots

Solve y'' - 5y' + 6y = 0.

Solution:

AE: m^2 - 5m + 6 = 0

(m - 2)(m - 3) = 0

m = 2 or m = 3

y = A e^(2x) + B e^(3x)

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4 Worked Example 2 — Repeated Root

Solve y'' - 4y' + 4y = 0.

Solution:

AE: m^2 - 4m + 4 = 0

(m - 2)^2 = 0

m = 2 (repeated)

y = (A + Bx) e^(2x)

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5 Worked Example 3 — Complex Roots

Solve y'' + 2y' + 5y = 0.

Solution:

AE: m^2 + 2m + 5 = 0

m = (-2 +/- sqrt(4 - 20)) / 2 = (-2 +/- sqrt(-16)) / 2 = -1 +/- 2i

So p = -1 and q = 2.

y = e^(-x) (A cos 2x + B sin 2x)

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6 Worked Example 4 — With Initial Conditions

Solve y'' - y' - 2y = 0, y(0) = 1, y'(0) = 4.

Solution:

AE: m^2 - m - 2 = 0, (m-2)(m+1) = 0, m = 2 or m = -1.

y = A e^(2x) + B e^(-x)

y' = 2A e^(2x) - B e^(-x)

From y(0) = 1: A + B = 1

From y'(0) = 4: 2A - B = 4

Adding: 3A = 5, so A = 5/3. Then B = 1 - 5/3 = -2/3.

y = (5/3) e^(2x) - (2/3) e^(-x)

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7 Non-Homogeneous Equations: Particular Integrals

For a y'' + b y' + c y = f(x), the general solution is:

y = y_CF + y_PI

where y_CF is the complementary function (general solution of the homogeneous equation) and y_PI is a particular integral (any one solution of the full equation).

Choosing the Form of the Particular Integral

f(x)	Try y_PI =
k (constant)	lambda (constant)
px + q	lambda x + mu
p e^(kx)	lambda e^(kx)
p sin kx or p cos kx	lambda cos kx + mu sin kx
Polynomial of degree n	Polynomial of degree n

Important Rule: If your trial y_PI is already part of the complementary function, multiply your trial by x (or x^2 if needed).

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8 Worked Example 5 — Exponential RHS

Solve y'' - 3y' + 2y = e^(4x).

Solution:

Step 1: CF. AE: m^2 - 3m + 2 = 0, (m-1)(m-2) = 0. y_CF = A e^x + B e^(2x).

Step 2: PI. Try y_PI = lambda e^(4x).

y_PI' = 4 lambda e^(4x), y_PI'' = 16 lambda e^(4x)

Substituting: 16 lambda - 12 lambda + 2 lambda = 1 (coefficient of e^(4x))

6 lambda = 1, lambda = 1/6.

y_PI = (1/6) e^(4x)

y = A e^x + B e^(2x) + (1/6) e^(4x)

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9 Worked Example 6 — Trigonometric RHS

Solve y'' + y = sin 2x.

Solution: