8.2.1 Solving Second Order Differential Equations

Introduction to Second-Order Differential Equations

Second-order differential equations (S.O.D.E.s) involve derivatives up to the second order. They are widely used in physics and engineering to model systems such as oscillations, damping, and forced motion.

General Form of a Homogeneous S.O.D.E.:

a \frac{d^2y}{dx^2} + b \frac{dy}{dx} + cy = 0

where $a$ , $b$ , and $c$ are constants.

The characteristic equation associated with the S.O.D.E. is:

am^2 + bm + c = 0

The solutions to the characteristic equation determine the behaviour of the system.

Classification of Roots and Solutions

Case 1: Distinct Real Roots ( $b^2 - 4ac > 0$ )

The characteristic equation has two distinct real roots, $m_1$ and $m_2$ .

Solution:

y = Ae^{m_1x} + Be^{m_2x}

where $A$ and $B$ are constants determined by initial conditions.

Case 2: Repeated Roots ( $b^2 - 4ac = 0$ )

The characteristic equation has a single repeated root, mm.

Solution:

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

Case 3: Complex Roots ( $b^2 - 4ac < 0$ )

The characteristic equation has complex roots m = $\alpha \pm \beta i$

Solution:

y = e^{\alpha x} \big(A\cos(\beta x) + B\sin(\beta x)\big)

Non-Homogeneous Second-Order Differential Equations

For equations of the form:

a \frac{d^2y}{dx^2} + b \frac{dy}{dx} + cy = F(x)

the general solution is:

y = y_c + y_p

where:

$y_c$ is the solution to the corresponding homogeneous equation.
$y_p$ is a particular integral found by guessing a form of $y_p$ based on $F(x)$ .

Worked Examples

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Example 1: Critical Damping

A particle of mass 0.5 kg moves in a straight line, subject to a restoring force proportional to displacement and a damping force proportional to velocity.

Solve:

$d2xdt2+8dxdt+16x=0$ , $\frac{d^2x}{dt^2} + 8\frac{dx}{dt} + 16x = 0$ ,

with initial conditions x(0)=1.5x(0) = 1.5 and $dxdt(0)=4\frac{dx}{dt}(0) = 4$ .

Step 1: Characteristic Equation

m^2 + 8m + 16 = 0 \quad \Rightarrow \quad (m + 4)^2 = 0

This is a repeated root ( $m = -4$ )

Step 2: General Solution

x = (A + Bt)e^{-4t}

Step 3: Apply Initial Conditions

At $t = 0, x = 1.5$

1.5 = (A + B \cdot 0)e^{0} \quad \Rightarrow \quad A = 1.5

At $t = 0, \frac{dx}{dt} = 4$

\frac{dx}{dt} = -4(A + Bt)e^{-4t} + Be^{-4t}

Substituting $t = 0$

4 = -4A + B \quad \Rightarrow \quad 4 = -4(1.5) + B \quad \Rightarrow \quad B = 10

Final Solution

x = (1.5 + 10t)e^{-4t}

infoNote

Example 2: Heavy Damping

Solve:

\frac{d^2x}{dt^2} + 8\frac{dx}{dt} + 12x = 0

with initial conditions $x(0) = 4$ and $\frac{dx}{dt}(0) = 0$

Step 1: Characteristic Equation

m^2 + 8m + 12 = 0 \quad \Rightarrow \quad (m + 6)(m + 2) = 0

The roots are $m=−6,−2$

Step 2: General Solution

x = Ae^{-6t} + Be^{-2t}

Step 3: Apply Initial Conditions

At $t = 0, x = 4$

4 = A + B

At $t = 0, \frac{dx}{dt} = 0$

\frac{dx}{dt} = -6Ae^{-6t} - 2Be^{-2t}

Substituting $t = 0$

0 = -6A - 2B

Step 4: Solve Simultaneous Equations

A + B = 4, \quad -6A - 2B = 0 \quad \Rightarrow \quad A = 6, B = -2

Final Solution

x = 6e^{-6t} - 2e^{-2t}

infoNote

Example 3: Light Damping

Solve:

\frac{d^2x}{dt^2} + 4\frac{dx}{dt} + 8x = 0,

with initial conditions $x(0) = 2$ and $\frac{dx}{dt}(0) = 0$

Step 1: Characteristic Equation

m^2 + 4m + 8 = 0 \quad \Rightarrow \quad m = -2 \pm 2i

Step 2: General Solution

x = e^{-2t}\big(A\cos(2t) + B\sin(2t)\big)

Step 3: Apply Initial Conditions

At $t = 0, x = 2$

2 = e^{0}(A \cdot 1 + B \cdot 0) \quad \Rightarrow \quad A = 2

At $t = 0, \frac{dx}{dt} = 0$

\frac{dx}{dt} = e^{-2t}\big(-2A\cos(2t) - 2B\sin(2t)\big) + e^{-2t}\big(-2A\sin(2t) + 2B\cos(2t)\big)

Substituting $t = 0$

0 = -2A + 2B \quad \Rightarrow \quad B = A = 2

Final Solution

x = e^{-2t}(2\cos(2t) + 2\sin(2t))

Note Summary

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Common Mistakes

Misidentifying the root type: Ensure correct classification as distinct, repeated, or complex roots.
Incorrect application of initial conditions: Verify substitutions carefully.
Forgetting the exponential factor for complex roots: Always include $e^{\alpha x}$ in the solution for complex roots.
Overlooking the particular integral in non-homogeneous cases.

infoNote

Key Formulas

Characteristic Equation:

am^2 + bm + c = 0

Distinct Real Roots ( $m_1, m_2$ )****:

y = Ae^{m_1x} + Be^{m_2x}

Repeated Roots ( $m$ ):

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

Complex Roots ( $\alpha \pm \beta i$ )****:

y = e^{\alpha x} \big(A\cos(\beta x) + B\sin(\beta x)\big)

Solving Second Order Differential Equations Simplified Revision Notes for A-Level Edexcel Further Maths Core Pure

8.2.1 Solving Second Order Differential Equations

Introduction to Second-Order Differential Equations

General Form of a Homogeneous S.O.D.E.:

Classification of Roots and Solutions

Case 1: Distinct Real Roots ( $b^2 - 4ac > 0$ )

Case 2: Repeated Roots ( $b^2 - 4ac = 0$ )

Case 3: Complex Roots ( $b^2 - 4ac < 0$ )

Non-Homogeneous Second-Order Differential Equations

Worked Examples

Example 1: Critical Damping

Example 2: Heavy Damping

Example 3: Light Damping

Note Summary

Common Mistakes

Key Formulas

500K+ Students Use These Powerful Tools to Master Solving Second Order Differential Equations For their A-Level Exams.

Other Revision Notes related to Solving Second Order Differential Equations you should explore

Solving Second Order Differential Equations Simplified Revision Notes for A-Level Edexcel Further Maths Core Pure

8.2.1 Solving Second Order Differential Equations

Introduction to Second-Order Differential Equations

General Form of a Homogeneous S.O.D.E.:

Classification of Roots and Solutions

Case 1: Distinct Real Roots (b2−4ac>0b^2 - 4ac > 0b2−4ac>0)

Case 2: Repeated Roots (b2−4ac=0b^2 - 4ac = 0b2−4ac=0)

Case 3: Complex Roots (b2−4ac<0b^2 - 4ac < 0b2−4ac<0)

Non-Homogeneous Second-Order Differential Equations

Worked Examples

Example 1: Critical Damping

Example 2**: Heavy Damping**

Example 3: Light Damping

Note Summary

Common Mistakes

Key Formulas

500K+ Students Use These Powerful Tools to Master Solving Second Order Differential Equations For their A-Level Exams.

Other Revision Notes related to Solving Second Order Differential Equations you should explore

Case 1: Distinct Real Roots ( $b^2 - 4ac > 0$ )

Case 2: Repeated Roots ( $b^2 - 4ac = 0$ )

Case 3: Complex Roots ( $b^2 - 4ac < 0$ )

Example 2: Heavy Damping