Solving Second Order Differential Equations (Edexcel A-Level Further Mathematics): Revision Notes

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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.:

ad2ydx2+bdydx+cy=0a \frac{d^2y}{dx^2} + b \frac{dy}{dx} + cy = 0

where aa, bb, and cc are constants.

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

am2+bm+c=0am^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 (b2−4ac>0b^2 - 4ac > 0)

The characteristic equation has two distinct real roots, m1m_1 and m2m_2.

Solution:

y=Aem1x+Bem2xy = Ae^{m_1x} + Be^{m_2x}

where AA and BB are constants determined by initial conditions.

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

The characteristic equation has a single repeated root, mm.

Solution:

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

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

The characteristic equation has complex roots m = α±βi\alpha \pm \beta i

Solution:

y=eαx(Acos⁡(βx)+Bsin⁡(βx))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:

ad2ydx2+bdydx+cy=F(x)a \frac{d^2y}{dx^2} + b \frac{dy}{dx} + cy = F(x)

the general solution is:

y=yc+ypy = y_c + y_p

where:

  • ycy_c is the solution to the corresponding homogeneous equation.
  • ypy_p is a particular integral found by guessing a form of ypy_p based on F(x)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=0d2xdt2+8dxdt+16x=0,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)=4dxdt(0)=4dxdt(0)=4\frac{dx}{dt}(0) = 4.

Step 1: Characteristic Equation

m2+8m+16=0⇒(m+4)2=0m^2 + 8m + 16 = 0 \quad \Rightarrow \quad (m + 4)^2 = 0

This is a repeated root (m=−4m = -4)

Step 2: General Solution

x=(A+Bt)e−4tx = (A + Bt)e^{-4t}

Step 3: Apply Initial Conditions

At t=0,x=1.5t = 0, x = 1.5

1.5=(A+B⋅0)e0⇒A=1.51.5 = (A + B \cdot 0)e^{0} \quad \Rightarrow \quad A = 1.5

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

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

Substituting t=0t = 0

4=−4A+B⇒4=−4(1.5)+B⇒B=104 = -4A + B \quad \Rightarrow \quad 4 = -4(1.5) + B \quad \Rightarrow \quad B = 10

Final Solution

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

Example 2**: Heavy Damping**

Solve:

d2xdt2+8dxdt+12x=0\frac{d^2x}{dt^2} + 8\frac{dx}{dt} + 12x = 0

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

Step 1: Characteristic Equation

m2+8m+12=0⇒(m+6)(m+2)=0m^2 + 8m + 12 = 0 \quad \Rightarrow \quad (m + 6)(m + 2) = 0

The roots are m=−6,−2m=−6,−2

Step 2: General Solution

x=Ae−6t+Be−2tx = Ae^{-6t} + Be^{-2t}

Step 3: Apply Initial Conditions

At t=0,x=4t = 0, x = 4

4=A+B4 = A + B

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

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

Substituting t=0t = 0

0=−6A−2B0 = -6A - 2B

Step 4: Solve Simultaneous Equations

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

Final Solution

x=6e−6t−2e−2tx = 6e^{-6t} - 2e^{-2t}
infoNote

Example 3: Light Damping

Solve:

d2xdt2+4dxdt+8x=0,\frac{d^2x}{dt^2} + 4\frac{dx}{dt} + 8x = 0,

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

Step 1: Characteristic Equation

m2+4m+8=0⇒m=−2±2im^2 + 4m + 8 = 0 \quad \Rightarrow \quad m = -2 \pm 2i

Step 2: General Solution

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

Step 3: Apply Initial Conditions

At t=0,x=2t = 0, x = 2

2=e0(A⋅1+B⋅0)⇒A=22 = e^{0}(A \cdot 1 + B \cdot 0) \quad \Rightarrow \quad A = 2

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

dxdt=e−2t(−2Acos⁡(2t)−2Bsin⁡(2t))+e−2t(−2Asin⁡(2t)+2Bcos⁡(2t))\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=0t = 0

0=−2A+2B⇒B=A=20 = -2A + 2B \quad \Rightarrow \quad B = A = 2

Final Solution

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

Note Summary

infoNote

Common Mistakes

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

Key Formulas

  1. Characteristic Equation:
am2+bm+c=0am^2 + bm + c = 0
  1. Distinct Real Roots (m1,m2m_1, m_2)****:
y=Aem1x+Bem2xy = Ae^{m_1x} + Be^{m_2x}
  1. Repeated Roots (mm):
y=(A+Bx)emxy = (A + Bx)e^{mx}
  1. Complex Roots (α±βi\alpha \pm \beta i)****:
y=eαx(Acos⁡(βx)+Bsin⁡(βx))y = e^{\alpha x} \big(A\cos(\beta x) + B\sin(\beta x)\big)

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