UNIT 4 • INITIAL VALUE PROBLEMS (ODE)

Taylor's Series
ODE Expansion Method

Taylor's Series Method computes exact localized polynomial approximations to initial value problems $y' = f(x, y)$ by repeatedly differentiating the differential equation at initial anchor point $(x_0, y_0)$ !

TAYLOR TAYLOR SERIES EXPANSIONDERIVATIVE STACKING

y(x) = y(x_0) + (x - x_0) y'(x_0) + \frac{(x - x_0)^2}{2!} y''(x_0) + \frac{(x - x_0)^3}{3!} y'''(x_0) + \cdots

1. Initial Base (y₀)Initial starting condition

y(x_0) = y_0

2. Successive DerivativesDifferentiate

y' = f(x, y)

repeatedly to find

y'', y''', y''''

3. Factorial DivisionEach term scales by

\frac{h^k}{k!}

where

h = x - x_0

Step-by-Step Worked Example

GUIDED STORY

OUR PROBLEM TO SOLVE

Solve the first-order differential equation:

\frac{dy}{dx} = x + y, \quad \text{with initial condition } y(0) = 1

Find the value of $y(0.1)$ using expansion up to the 4th derivative!

STEP 01 • EVALUATE HIGHER-ORDER DERIVATIVES AT (0, 1)

Differentiate $y' = x + y$ repeatedly with respect to $x$ :

y'(0) = 0 + 1 = 1

y''(x) = 1 + y' \implies y''(0) = 1 + 1 = 2

y'''(x) = y'' \implies y'''(0) = 2

y''''(x) = y''' \implies y''''(0) = 2

STEP 02 • NUMERICAL TAYLOR EXPANSION

Plug derivatives into Taylor series formula for $h = 0.1$ :

y(0.1) = y_0 + h y'_0 + \frac{h^2}{2!} y''_0 + \frac{h^3}{3!} y'''_0 + \frac{h^4}{4!} y^{(4)}_0

y(0.1) = 1.00000 + (0.1)(1) + \frac{(0.1)^2}{2}(2) + \frac{(0.1)^3}{6}(2) + \frac{(0.1)^4}{24}(2)

y(0.1) = 1.00000 + 0.10000 + 0.01000 + 0.000333 + 0.000008

y(0.1) = 1.11034

FINAL RESULT

Approximated Value: y(0.1) ≈ 1.11034

\boxed{y(0.1) \approx 1.11034} \quad (\text{Exact: } 2e^{0.1} - 1.1 \approx 1.11034)

Our 4th-order Taylor series approximation 1.11034 matches the exact analytical solution $y(x) = 2e^x - x - 1$ to 5 decimal places!

LABORATORY

Interactive Live Calculator & Visualizer

SERIES EXPANSION ODE LABORATORY

Taylor Series Expansion Engine

PreviousNumerical Differentiation

Unit 4•18 of 28

NextEuler's Method

∑

f(x)∫dxπ

NETZ Engine

Evaluating numerical algorithm

UNIT 4 • INITIAL VALUE PROBLEMS (ODE)

Taylor's Series
ODE Expansion Method

TAYLOR TAYLOR SERIES EXPANSIONDERIVATIVE STACKING

y(x) = y(x_0) + (x - x_0) y'(x_0) + \frac{(x - x_0)^2}{2!} y''(x_0) + \frac{(x - x_0)^3}{3!} y'''(x_0) + \cdots

1. Initial Base (y₀)Initial starting condition

y(x_0) = y_0

2. Successive DerivativesDifferentiate

y' = f(x, y)

repeatedly to find

y'', y''', y''''

3. Factorial DivisionEach term scales by

\frac{h^k}{k!}

where

h = x - x_0

Step-by-Step Worked Example

GUIDED STORY

OUR PROBLEM TO SOLVE

Solve the first-order differential equation:

\frac{dy}{dx} = x + y, \quad \text{with initial condition } y(0) = 1

Find the value of $y(0.1)$ using expansion up to the 4th derivative!

STEP 01 • EVALUATE HIGHER-ORDER DERIVATIVES AT (0, 1)

Differentiate $y' = x + y$ repeatedly with respect to $x$ :

y'(0) = 0 + 1 = 1

y''(x) = 1 + y' \implies y''(0) = 1 + 1 = 2

y'''(x) = y'' \implies y'''(0) = 2

y''''(x) = y''' \implies y''''(0) = 2

STEP 02 • NUMERICAL TAYLOR EXPANSION

Plug derivatives into Taylor series formula for $h = 0.1$ :

y(0.1) = y_0 + h y'_0 + \frac{h^2}{2!} y''_0 + \frac{h^3}{3!} y'''_0 + \frac{h^4}{4!} y^{(4)}_0

y(0.1) = 1.00000 + (0.1)(1) + \frac{(0.1)^2}{2}(2) + \frac{(0.1)^3}{6}(2) + \frac{(0.1)^4}{24}(2)

y(0.1) = 1.00000 + 0.10000 + 0.01000 + 0.000333 + 0.000008

y(0.1) = 1.11034

FINAL RESULT

Approximated Value: y(0.1) ≈ 1.11034

\boxed{y(0.1) \approx 1.11034} \quad (\text{Exact: } 2e^{0.1} - 1.1 \approx 1.11034)

Our 4th-order Taylor series approximation 1.11034 matches the exact analytical solution $y(x) = 2e^x - x - 1$ to 5 decimal places!

LABORATORY

Interactive Live Calculator & Visualizer

SERIES EXPANSION ODE LABORATORY

Taylor Series Expansion Engine

PreviousNumerical Differentiation

Unit 4•18 of 28

NextEuler's Method

Taylor's Series ODE Expansion Method

Step-by-Step Worked Example

Approximated Value: y(0.1) ≈ 1.11034

Interactive Live Calculator & Visualizer

Taylor Series Expansion Engine

NETZ Engine

Taylor's Series ODE Expansion Method

Step-by-Step Worked Example

Approximated Value: y(0.1) ≈ 1.11034

Interactive Live Calculator & Visualizer

Taylor Series Expansion Engine

Taylor's Series
ODE Expansion Method

Taylor's Series
ODE Expansion Method