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3.3 Ordinary Differential Equations & Initial-Value Problems
PRC Registered Electrical Engineer REE
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3.3 Ordinary Differential Equations & Initial-Value Problems
Key Takeaways Explain This Topic AI
Ordinary differential equations (ODEs) categorize dynamic physical systems by order, degree, and linearity, modeling electrical circuit transient behavior.
First-order ODE solution methods include Separation of Variables, Exactness testing, Integrating Factors ( I ( x ) = e ∫ P d x I(x) = e^{\int P \, dx} I ( x ) = e ∫ P d x ), and Bernoulli substitutions.
Second-order linear ODEs with constant coefficients a y ′ ′ + b y ′ + c y = 0 a y'' + b y' + c y = 0 a y ′′ + b y ′ + cy = 0 yield overdamped, critically damped, or underdamped responses determined by the characteristic discriminant.
Non-homogeneous ODEs ( a y ′ ′ + b y ′ + c y = g ( t ) a y'' + b y' + c y = g(t) a y ′′ + b y ′ + cy = g ( t ) ) have general solution y ( t ) = y h ( t ) + y p ( t ) y(t) = y_h(t) + y_p(t) y ( t ) = y h ( t ) + y p ( t ) , solved via Undetermined Coefficients or Variation of Parameters.
In electrical circuits, time constants τ = L / R \tau = L/R τ = L / R (RL circuits) and τ = R C \tau = RC τ = RC (RC circuits) govern exponential growth and decay rates of transient current and voltage.
Last updated: August 2026
3.3 Ordinary Differential Equations & Initial-Value Problems
Ordinary Differential Equations (ODEs) form the dynamic mathematical backbone of electrical engineering. Transient phenomena in electrical circuits containing energy storage elements (inductors and capacitors), control systems, electrical machinery dynamics, and electromagnetic waves are modeled by ordinary differential equations.
1. Classification of Differential Equations
Order: The order of the highest derivative present in the equation.
Degree: The algebraic exponent of the highest-order derivative after removing fractional exponents.
Linearity: An ODE is linear if the dependent variable y y y and its derivatives appear only to the first power and are not multiplied together (e.g., a n ( x ) y ( n ) + ⋯ + a 1 ( x ) y ′ + a 0 ( x ) y = g ( x ) a_n(x) y^{(n)} + \dots + a_1(x) y' + a_0(x) y = g(x) a n ( x ) y ( n ) + ⋯ + a 1 ( x ) y ′ + a 0 ( x ) y = g ( x ) ).
Homogeneity: Homogeneous if g ( x ) = 0 g(x) = 0 g ( x ) = 0 ; non-homogeneous if g ( x ) ≠ 0 g(x) \neq 0 g ( x ) = 0 .
- 2. First-Order Differential Equations
- 1. Variable Separable ODEs
Form: d y d x = g ( x ) h ( y ) ⟹ ∫ d y h ( y ) = ∫ g ( x ) d x + C \frac{dy}{dx} = g(x) h(y) \implies \int \frac{dy}{h(y)} = \int g(x) \, dx + C d x d y = g ( x ) h ( y ) ⟹ ∫ h ( y ) d y = ∫ g ( x ) d x + C
- 2. Exact Differential Equations
- Form: M ( x , y ) d x + N ( x , y ) d y = 0 M(x,y)\,dx + N(x,y)\,dy = 0 M ( x , y ) d x + N ( x , y ) d y = 0
Test for Exactness: ∂ M ∂ y = ∂ N ∂ x \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} ∂ y ∂ M = ∂ x ∂ N
Solution is given by potential function Φ ( x , y ) = C \Phi(x,y) = C Φ ( x , y ) = C where ∂ Φ ∂ x = M \frac{\partial \Phi}{\partial x} = M ∂ x ∂ Φ = M and ∂ Φ ∂ y = N \frac{\partial \Phi}{\partial y} = N ∂ y ∂ Φ = N .
If not exact, an Integrating Factor μ ( x ) = e ∫ M y − N x N d x \mu(x) = e^{\int \frac{M_y - N_x}{N} \, dx} μ ( x ) = e ∫ N M y − N x d x converts it to an exact ODE.
- 3. First-Order Linear ODEs
- Standard Form: d y d x + P ( x ) y = Q ( x ) \frac{dy}{dx} + P(x)y = Q(x) d x d y + P ( x ) y = Q ( x )
- Integrating Factor: I ( x ) = e ∫ P ( x ) d x I(x) = e^{\int P(x) \, dx} I ( x ) = e ∫ P ( x ) d x
General Solution: y ⋅ I ( x ) = ∫ Q ( x ) I ( x ) d x + C ⟹ y ( x ) = e − ∫ P d x [ ∫ Q ( x ) e ∫ P d x d x + C ] y \cdot I(x) = \int Q(x) I(x) \, dx + C \implies y(x) = e^{-\int P dx} \left[ \int Q(x) e^{\int P dx} \, dx + C \right] y ⋅ I ( x ) = ∫ Q ( x ) I ( x ) d x + C ⟹ y ( x ) = e − ∫ P d x [ ∫ Q ( x ) e ∫ P d x d x + C ]
4. Bernoulli Equations
Form: d y d x + P ( x ) y = Q ( x ) y n ( n ≠ 0 , 1 ) \frac{dy}{dx} + P(x)y = Q(x)y^n \quad (n \neq 0, 1) d x d y + P ( x ) y = Q ( x ) y n ( n = 0 , 1 )
Substitution u = y 1 − n u = y^{1-n} u = y 1 − n transforms the non-linear equation into a linear ODE in terms of u ( x ) u(x) u ( x ) : d u d x + ( 1 − n ) P ( x ) u = ( 1 − n ) Q ( x ) \frac{du}{dx} + (1-n) P(x) u = (1-n) Q(x) d x d u + ( 1 − n ) P ( x ) u = ( 1 − n ) Q ( x )
3. Second-Order Linear Homogeneous ODEs with Constant Coefficients
Standard Form: a d 2 y d t 2 + b d y d t + c y = 0 a \frac{d^2y}{dt^2} + b \frac{dy}{dt} + c y = 0 a d t 2 d 2 y + b d t d y + cy = 0
Characteristic (Auxiliary) Equation
Substituting y = e r t y = e^{rt} y = e r t yields: a r 2 + b r + c = 0 ⟹ r 1 , 2 = − b ± b 2 − 4 a c 2 a a r^2 + b r + c = 0 \implies r_{1,2} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} a r 2 + b r + c = 0 ⟹ r 1 , 2 = 2 a − b ± b 2 − 4 a c
Depending on the discriminant Δ = b 2 − 4 a c \Delta = b^2 - 4ac Δ = b 2 − 4 a c , three distinct physical damping regimes arise:
Discriminant Δ \Delta Δ Root Characteristics Solution Form y h ( t ) y_h(t) y h ( t ) Physical Transient Response
b 2 − 4 a c > 0 b^2 - 4ac > 0 b 2 − 4 a c > 0 Real & Distinct ( r 1 ≠ r 2 r_1 \neq r_2 r 1 = r 2 ) y h ( t ) = C 1 e r 1 t + C 2 e r 2 t y_h(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t} y h ( t ) = C 1 e r 1 t + C 2 e r 2 t Overdamped (Slow decay, no oscillation)
b 2 − 4 a c = 0 b^2 - 4ac = 0 b 2 − 4 a c = 0 Real & Repeated ( r 1 = r 2 = r r_1 = r_2 = r r 1 = r 2 = r ) y h ( t ) = ( C 1 + C 2 t ) e r t y_h(t) = (C_1 + C_2 t) e^{r t} y h ( t ) = ( C 1 + C 2 t ) e r t Critically Damped (Fastest decay without overshoot)
b 2 − 4 a c < 0 b^2 - 4ac < 0 b 2 − 4 a c < 0 Complex Conjugate ( r = α ± j β r = \alpha \pm j\beta r = α ± j β ) y h ( t ) = e α t ( C 1 cos β t + C 2 sin β t ) y_h(t) = e^{\alpha t} (C_1 \cos \beta t + C_2 \sin \beta t) y h ( t ) = e α t ( C 1 cos βt + C 2 sin βt ) Underdamped (Decaying sinusoidal oscillations)
Where α = − b 2 a \alpha = -\frac{b}{2a} α = − 2 a b (attenuation constant) and β = 4 a c − b 2 2 a \beta = \frac{\sqrt{4ac - b^2}}{2a} β = 2 a 4 a c − b 2 (damped angular frequency).
4. Second-Order Non-Homogeneous ODEs
Form: a y ′ ′ + b y ′ + c y = g ( t ) a y'' + b y' + c y = g(t) a y ′′ + b y ′ + cy = g ( t ) General Solution: y ( t ) = y h ( t ) + y p ( t ) y(t) = y_h(t) + y_p(t) y ( t ) = y h ( t ) + y p ( t ) (Natural Transient + Forced Steady-State Solution).
Method of Undetermined Coefficients
Used when forcing function g ( t ) g(t) g ( t ) consists of polynomials, exponentials, sines, or cosines:
- Forcing Function g ( t ) g(t) g ( t ) Trial Particular Solution y p ( t ) y_p(t) y p ( t )
- k e γ t k e^{\gamma t} k e γ t A e γ t A e^{\gamma t} A e γ t
k n t n + ⋯ + k 0 k_n t^n + \dots + k_0 k n t n + ⋯ + k 0 A n t n + A n − 1 t n − 1 + ⋯ + A 0 A_n t^n + A_{n-1} t^{n-1} + \dots + A_0 A n t n + A n − 1 t n − 1 + ⋯ + A 0
k cos ( ω t ) k \cos(\omega t) k cos ( ω t ) or k sin ( ω t ) k \sin(\omega t) k sin ( ω t ) A cos ( ω t ) + B sin ( ω t ) A \cos(\omega t) + B \sin(\omega t) A cos ( ω t ) + B sin ( ω t )
Note: If any term in the trial y p ( t ) y_p(t) y p ( t ) duplicates a term in y h ( t ) y_h(t) y h ( t ) , multiply the trial solution by t t t (or t 2 t^2 t 2 ).
Variation of Parameters
Universal method applicable for any continuous forcing function g ( t ) g(t) g ( t ) : y p ( t ) = − y 1 ( t ) ∫ y 2 ( t ) g ( t ) W ( t ) d t + y 2 ( t ) ∫ y 1 ( t ) g ( t ) W ( t ) d t y_p(t) = -y_1(t) \int \frac{y_2(t) g(t)}{W(t)} \, dt + y_2(t) \int \frac{y_1(t) g(t)}{W(t)} \, dt y p ( t ) = − y 1 ( t ) ∫ W ( t ) y 2 ( t ) g ( t ) d t + y 2 ( t ) ∫ W ( t ) y 1 ( t ) g ( t ) d t
where the Wronskian is W ( t ) = y 1 y 2 ′ − y 1 ′ y 2 W(t) = y_1 y_2' - y_1' y_2 W ( t ) = y 1 y 2 ′ − y 1 ′ y 2 .
- 5. Electrical Circuit Applications (Initial-Value Problems)
- Series RL Circuit
Governed by Kirchhoff's Voltage Law (KVL): L d i d t + R i = V ( t ) L \frac{di}{dt} + R i = V(t) L d t d i + R i = V ( t )
For a step DC voltage input V ( t ) = V 0 V(t) = V_0 V ( t ) = V 0 at t = 0 t = 0 t = 0 with i ( 0 ) = 0 i(0) = 0 i ( 0 ) = 0 : ∋ ( t ) = V 0 R ( 1 − e − t / τ ) where τ = L R (Time Constant) \ni(t) = \frac{V_0}{R} \left(1 - e^{-t/\tau}\right) \quad \text{where } \tau = \frac{L}{R} \text{ (Time Constant)} ∋ ( t ) = R V 0 ( 1 − e − t / τ ) where τ = R L (Time Constant)
Series RC Circuit
Governed by KVL in terms of charge q ( t ) q(t) q ( t ) : R d q d t + 1 C q = V ( t ) ⟹ R d i d t + 1 C i = d V d t R \frac{dq}{dt} + \frac{1}{C} q = V(t) \implies R \frac{di}{dt} + \frac{1}{C} i = \frac{dV}{dt} R d t d q + C 1 q = V ( t ) ⟹ R d t d i + C 1 i = d t d V
For a step DC voltage charging a capacitor with initial charge q ( 0 ) = 0 q(0) = 0 q ( 0 ) = 0 : ∋ ( t ) = V 0 R e − t / τ where τ = R C (Time Constant) \ni(t) = \frac{V_0}{R} e^{-t/\tau} \quad \text{where } \tau = R C \text{ (Time Constant)} ∋ ( t ) = R V 0 e − t / τ where τ = RC (Time Constant)
Series RLC Circuit
KVL formulation: L d 2 q d t 2 + R d q d t + 1 C q = E ( t ) ⟹ L d 2 i d t 2 + R d i d t + 1 C i = d E d t L \frac{d^2q}{dt^2} + R \frac{dq}{dt} + \frac{1}{C} q = E(t) \implies L \frac{d^2i}{dt^2} + R \frac{di}{dt} + \frac{1}{C} i = \frac{dE}{dt} L d t 2 d 2 q + R d t d q + C 1 q = E ( t ) ⟹ L d t 2 d 2 i + R d t d i + C 1 i = d t d E
Key circuit metrics:
- Undamped natural frequency: ω 0 = 1 L C \omega_0 = \frac{1}{\sqrt{L C}} ω 0 = L C 1
- Damping ratio: ζ = R 2 C L \zeta = \frac{R}{2} \sqrt{\frac{C}{L}} ζ = 2 R L C
- ζ > 1 ⟹ \zeta > 1 \implies ζ > 1 ⟹ Overdamped
- ζ = 1 ⟹ \zeta = 1 \implies ζ = 1 ⟹ Critically Damped
- ζ < 1 ⟹ \zeta < 1 \implies ζ < 1 ⟹ Underdamped
- Step-by-Step Worked Numerical Examples
- Example 3.3.1: First-Order RL Circuit Transient IVP
Problem: A series RL circuit has resistance R = 10 Ω R = 10\ \Omega R = 10 Ω , inductance L = 2 H L = 2\text{ H} L = 2 H , and a DC voltage source V = 50 V V = 50\text{ V} V = 50 V connected at t = 0 t = 0 t = 0 . Given initial current i ( 0 ) = 0 A i(0) = 0\text{ A} i ( 0 ) = 0 A , find i ( t ) i(t) i ( t ) and calculate the current at t = 0.4 seconds t = 0.4\text{ seconds} t = 0.4 seconds .
Solution:
Formulate ODE: 2 d i d t + 10 i = 50 ⟹ d i d t + 5 i = 25 2 \frac{di}{dt} + 10 i = 50 \implies \frac{di}{dt} + 5 i = 25 2 d t d i + 10 i = 50 ⟹ d t d i + 5 i = 25
Find Integrating Factor I ( t ) I(t) I ( t ) : P ( t ) = 5 ⟹ I ( t ) = e ∫ 5 d t = e 5 t P(t) = 5 \implies I(t) = e^{\int 5 \, dt} = e^{5t} P ( t ) = 5 ⟹ I ( t ) = e ∫ 5 d t = e 5 t
Multiply ODE by I ( t ) I(t) I ( t ) and integrate: d d t ( i e 5 t ) = 25 e 5 t ⟹ i e 5 t = ∫ 25 e 5 t d t = 5 e 5 t + C \frac{d}{dt}\left(i e^{5t}\right) = 25 e^{5t} \implies i e^{5t} = \int 25 e^{5t} \, dt = 5 e^{5t} + C d t d ( i e 5 t ) = 25 e 5 t ⟹ i e 5 t = ∫ 25 e 5 t d t = 5 e 5 t + C i ( t ) = 5 + C e − 5 t i(t) = 5 + C e^{-5t} i ( t ) = 5 + C e − 5 t
Apply initial condition i ( 0 ) = 0 i(0) = 0 i ( 0 ) = 0 : 0 = 5 + C ( 1 ) ⟹ C = − 5 0 = 5 + C(1) \implies C = -5 0 = 5 + C ( 1 ) ⟹ C = − 5 i ( t ) = 5 ( 1 − e − 5 t ) A i(t) = 5 \left(1 - e^{-5t}\right)\text{ A} i ( t ) = 5 ( 1 − e − 5 t ) A
Evaluate at t = 0.4 s t = 0.4\text{ s} t = 0.4 s : i ( 0.4 ) = 5 ( 1 − e − 5 ( 0.4 ) ) = 5 ( 1 − e − 2 ) = 5 ( 1 − 0.135335 ) = 5 ( 0.864665 ) = 4.3233 A i(0.4) = 5 \left(1 - e^{-5(0.4)}\right) = 5 \left(1 - e^{-2}\right) = 5 (1 - 0.135335) = 5(0.864665) = 4.3233\text{ A} i ( 0.4 ) = 5 ( 1 − e − 5 ( 0.4 ) ) = 5 ( 1 − e − 2 ) = 5 ( 1 − 0.135335 ) = 5 ( 0.864665 ) = 4.3233 A
Example 3.3.2: RLC Circuit Transient Damping Regime Analysis
Problem: A series RLC circuit has L = 1 H L = 1\text{ H} L = 1 H and C = 0.25 F C = 0.25\text{ F} C = 0.25 F . Determine the response regime (overdamped, critically damped, or underdamped) and the natural response q h ( t ) q_h(t) q h ( t ) for R = 5 Ω R = 5\ \Omega R = 5 Ω .
Solution:
Formulate characteristic equation: L r 2 + R r + 1 C = 0 ⟹ 1 r 2 + 5 r + 1 0.25 = 0 ⟹ r 2 + 5 r + 4 = 0 L r^2 + R r + \frac{1}{C} = 0 \implies 1 r^2 + 5 r + \frac{1}{0.25} = 0 \implies r^2 + 5r + 4 = 0 L r 2 + R r + C 1 = 0 ⟹ 1 r 2 + 5 r + 0.25 1 = 0 ⟹ r 2 + 5 r + 4 = 0
Find roots of characteristic equation: ( r + 1 ) ( r + 4 ) = 0 ⟹ r 1 = − 1 , r 2 = − 4 (r + 1)(r + 4) = 0 \implies r_1 = -1, \quad r_2 = -4 ( r + 1 ) ( r + 4 ) = 0 ⟹ r 1 = − 1 , r 2 = − 4
Identify damping regime: Since r 1 , r 2 r_1, r_2 r 1 , r 2 are real and distinct (discriminant Δ = 5 2 − 4 ( 1 ) ( 4 ) = 9 > 0 \Delta = 5^2 - 4(1)(4) = 9 > 0 Δ = 5 2 − 4 ( 1 ) ( 4 ) = 9 > 0 ), the circuit response is Overdamped .
Write natural transient charge solution: q h ( t ) = C 1 e − t + C 2 e − 4 t q_h(t) = C_1 e^{-t} + C_2 e^{-4t} q h ( t ) = C 1 e − t + C 2 e − 4 t
Example 3.3.3: Second-Order Non-Homogeneous Resonant System
Problem: Solve the initial value problem y ′ ′ + 4 y = 12 cos ( 2 t ) y'' + 4y = 12 \cos(2t) y ′′ + 4 y = 12 cos ( 2 t ) with y ( 0 ) = 0 y(0) = 0 y ( 0 ) = 0 and y ′ ( 0 ) = 0 y'(0) = 0 y ′ ( 0 ) = 0 .
Solution:
Homogeneous solution y h ( t ) y_h(t) y h ( t ) : r 2 + 4 = 0 ⟹ r = ± j 2 ⟹ y h ( t ) = C 1 cos ( 2 t ) + C 2 sin ( 2 t ) r^2 + 4 = 0 \implies r = \pm j2 \implies y_h(t) = C_1 \cos(2t) + C_2 \sin(2t) r 2 + 4 = 0 ⟹ r = ± j 2 ⟹ y h ( t ) = C 1 cos ( 2 t ) + C 2 sin ( 2 t )
Formulate trial particular solution y p ( t ) y_p(t) y p ( t ) : Because forcing term 12 cos ( 2 t ) 12 \cos(2t) 12 cos ( 2 t ) duplicates terms in y h ( t ) y_h(t) y h ( t ) , multiply trial by t t t : y p ( t ) = t ( A cos 2 t + B sin 2 t ) y_p(t) = t (A \cos 2t + B \sin 2t) y p ( t ) = t ( A cos 2 t + B sin 2 t )
Differentiate y p ( t ) y_p(t) y p ( t ) : y p ′ = ( A cos 2 t + B sin 2 t ) + t ( − 2 A sin 2 t + 2 B cos 2 t ) y_p' = (A \cos 2t + B \sin 2t) + t (-2A \sin 2t + 2B \cos 2t) y p ′ = ( A cos 2 t + B sin 2 t ) + t ( − 2 A sin 2 t + 2 B cos 2 t ) y p ′ ′ = − 4 A sin 2 t + 4 B cos 2 t + t ( − 4 A cos 2 t − 4 B sin 2 t ) y_p'' = -4A \sin 2t + 4B \cos 2t + t (-4A \cos 2t - 4B \sin 2t) y p ′′ = − 4 A sin 2 t + 4 B cos 2 t + t ( − 4 A cos 2 t − 4 B sin 2 t )
Substitute into ODE y p ′ ′ + 4 y p = 12 cos 2 t y_p'' + 4y_p = 12 \cos 2t y p ′′ + 4 y p = 12 cos 2 t : − 4 A sin 2 t + 4 B cos 2 t − 4 y p + 4 y p = 12 cos 2 t -4A \sin 2t + 4B \cos 2t - 4 y_p + 4 y_p = 12 \cos 2t − 4 A sin 2 t + 4 B cos 2 t − 4 y p + 4 y p = 12 cos 2 t − 4 A sin 2 t + 4 B cos 2 t = 12 cos 2 t -4A \sin 2t + 4B \cos 2t = 12 \cos 2t − 4 A sin 2 t + 4 B cos 2 t = 12 cos 2 t Equating coefficients: − 4 A = 0 ⟹ A = 0 -4A = 0 \implies A = 0 − 4 A = 0 ⟹ A = 0 , and 4 B = 12 ⟹ B = 3 4B = 12 \implies B = 3 4 B = 12 ⟹ B = 3 . y p ( t ) = 3 t sin ( 2 t ) y_p(t) = 3t \sin(2t) y p ( t ) = 3 t sin ( 2 t )
Apply initial conditions to general solution y ( t ) = C 1 cos 2 t + C 2 sin 2 t + 3 t sin 2 t y(t) = C_1 \cos 2t + C_2 \sin 2t + 3t \sin 2t y ( t ) = C 1 cos 2 t + C 2 sin 2 t + 3 t sin 2 t :
y ( 0 ) = C 1 + 0 + 0 = 0 ⟹ C 1 = 0 y(0) = C_1 + 0 + 0 = 0 \implies C_1 = 0 y ( 0 ) = C 1 + 0 + 0 = 0 ⟹ C 1 = 0
y ′ ( t ) = 2 C 2 cos 2 t + 3 sin 2 t + 6 t cos 2 t y'(t) = 2C_2 \cos 2t + 3 \sin 2t + 6t \cos 2t y ′ ( t ) = 2 C 2 cos 2 t + 3 sin 2 t + 6 t cos 2 t
y ′ ( 0 ) = 2 C 2 + 0 + 0 = 0 ⟹ C 2 = 0 y'(0) = 2C_2 + 0 + 0 = 0 \implies C_2 = 0 y ′ ( 0 ) = 2 C 2 + 0 + 0 = 0 ⟹ C 2 = 0