Taylor's Series Method for Ordinary Differential Equations

Taylor's Series Method is a numerical technique used to find an approximate solution of an ordinary differential equation (ODE) by expanding the solution as a Taylor series around a known initial point.

It is particularly useful when the differential equation and its successive higher-order derivatives can be evaluated easily at the initial point.

Formula for Taylor Series Expansion:

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

y(x) \approx y_0 + (x - x_0) y'_0 + \frac{(x - x_0)^2}{2!} y''_0 + \dots + \frac{(x - x_0)^n}{n!} y^{(n)}_0

Where:

$x_0$ is the initial value of $x$ .
$y_0 = y(x_0)$ is the initial value of $y$ .
$y'_0, y''_0, \dots$ are successive derivatives at $x = x_0$ .

Example of Taylor's Series Method

Solve Differential Equation

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

Find $y(0.1)$ up to 4th derivative

Step 1 to 4: Compute Derivatives at $x_0 = 0$

y'(0) = 0 + 1 = 1

y''(0) = 1 + y'(0) = 2

y'''(0) = y''(0) = 2

y''''(0) = y'''(0) = 2

Step 5 to 7: Apply Expansion for $x = 0.1$

y(0.1) = 1 + (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 + 0.1 + 0.01 + 0.000333 + 0.000008 = 1.11034

Conclusion

\boxed{y(0.1) \approx 1.11034}

Therefore, using Taylor's Series Method up to the fourth derivative, the value of $y$ at $x = 0.1$ is approximately 1.11034.

Interactive Taylor's Series Calculator

Taylor's Series ODE Solver

PreviousNumerical Differentiation

Unit 4•18 of 28

NextEuler's Method

Taylor's Series Method for Ordinary Differential Equations

It is particularly useful when the differential equation and its successive higher-order derivatives can be evaluated easily at the initial point.

Formula for Taylor Series Expansion:

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

y(x) \approx y_0 + (x - x_0) y'_0 + \frac{(x - x_0)^2}{2!} y''_0 + \dots + \frac{(x - x_0)^n}{n!} y^{(n)}_0

Where:

$x_0$ is the initial value of $x$ .
$y_0 = y(x_0)$ is the initial value of $y$ .
$y'_0, y''_0, \dots$ are successive derivatives at $x = x_0$ .

Example of Taylor's Series Method

Solve Differential Equation

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

Find $y(0.1)$ up to 4th derivative

Step 1 to 4: Compute Derivatives at $x_0 = 0$

y'(0) = 0 + 1 = 1

y''(0) = 1 + y'(0) = 2

y'''(0) = y''(0) = 2

y''''(0) = y'''(0) = 2

Step 5 to 7: Apply Expansion for $x = 0.1$

y(0.1) = 1 + (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 + 0.1 + 0.01 + 0.000333 + 0.000008 = 1.11034

Conclusion

\boxed{y(0.1) \approx 1.11034}

Therefore, using Taylor's Series Method up to the fourth derivative, the value of $y$ at $x = 0.1$ is approximately 1.11034.

Interactive Taylor's Series Calculator

Taylor's Series ODE Solver

PreviousNumerical Differentiation

Unit 4•18 of 28

NextEuler's Method

Taylor's Series Method for Ordinary Differential Equations

Example of Taylor's Series Method

Step 1 to 4: Compute Derivatives at x0=0x_0 = 0x0​=0

Step 5 to 7: Apply Expansion for x=0.1x = 0.1x=0.1

Conclusion

Interactive Taylor's Series Calculator

Taylor's Series ODE Solver

NETZ Engine

Taylor's Series Method for Ordinary Differential Equations

Example of Taylor's Series Method

Step 1 to 4: Compute Derivatives at x0=0x_0 = 0x0​=0

Step 5 to 7: Apply Expansion for x=0.1x = 0.1x=0.1

Conclusion

Interactive Taylor's Series Calculator

Taylor's Series ODE Solver

Step 1 to 4: Compute Derivatives at $x_0 = 0$

Step 5 to 7: Apply Expansion for $x = 0.1$

Step 1 to 4: Compute Derivatives at $x_0 = 0$

Step 5 to 7: Apply Expansion for $x = 0.1$