Eulers Method Practice Problems

Euler method is a numerical technique used to approximate solutions to ordinary differential equations (ODEs). It is particularly useful when exact solutions are difficult or impossible to find. The method is named after the Swiss mathematician Leonhard Euler, who developed it in the 18th century.

In this article, we will understand what Euler's method is, and solve problems related to it.

What is Euler's Method?

The Euler method is a fundamental numerical technique used for approximating solutions to ordinary differential equations (ODEs). It is one of the simplest and oldest methods for solving initial value problems where an exact analytical solution might be difficult or impossible.

The Euler method provides an iterative process to approximate the solution of a first-order ODE of the form:

\frac{dy}{dt} = f(t, y)

with an initial condition y(t₀) = y₀.

Example of Euler method

Consider the differential equation:

\frac{dy}{dt} = y

with the initial condition:

y(0)=1

We want to approximate the solution at t=0.2 using the Euler method with a step size h=0.1.

Important formulas

y_n+1​ = y_n​+h⋅f(t_n​,y_n​)

where:

• y_n+1is the approximated value of y at t_n+1​,
• y_n​ is the value of y at tnt_ntn​,
• h is the step size,
• f(t_n, y_n) is the derivative \frac{dy}{dt} evaluated at (t_n, y_n).

Step Size: The step size h determines the increment in ttt for each step and is typically chosen based on desired accuracy and stability.

For a system of ODEs, such as:

\frac{dx}{dt} = f(t, x, y)

\frac{dy}{dt} = g(t, x, y)

The Euler method is applied to each equation individually:

x_{n+1} = x_n + h \cdot f(t_n, x_n, y_n)

y_{n+1} = y_n + h \cdot g(t_n, x_n, y_n)Euler's

General Steps

The general steps include:

1. Initialisation: Set t₀, y₀​, and choose a step size h.
2. Iteration: For each step n from 0 to N−1 (where t_N=t₀+N⋅h)

=t_{n+1} = t_n + h

=y_{n+1} = y_n + h \cdot f(t_n, y_n)

3. Error Analysis: Note that the Euler method introduces truncation errors, and its accuracy depends on the step size hhh. Smaller hhh generally results in more accurate results but requires more computations.

Eulers Method: Practice Problems with solutions

P1. ODE: \frac{dy}{dt} = y - t^2 + 1

Initial Condition: y(0)=0.5

Step Size (h): 0.2

Solution:

Calculate the value at t=0.2:

y_{n+1} = y_n + h \cdot f(t_n, y_n)

Where f(t,y) = y - t^2 + 1. For t₀=0 and y₀=0.5:

f(0,0.5)=0.5−0²+1=1.5

So,

y_1 = y_0 + h \cdot f(0, 0.5) = 0.5 + 0.2 \cdot 1.5 = 0.5 + 0.3 = 0.8

Thus, y(0.2) \approx 0.8.

Calculate the value at t=0.4:

For t₁​=0.2 and y₁=0.8:

f(0.2,0.8)=0.8−0.2²+1=0.8−0.04+1=1.76

y_2 = y_1 + h \cdot f(0.2, 0.8) = 0.8 + 0.2 \cdot 1.76 = 0.8 + 0.352 = 1.152

Thus, y(0.4) \approx 1.152.

P2. ODE: \frac{dy}{dt} = y \cdot \sin(t)

Initial Condition: y(0)=1

Step Size (h): 0.1

Solution:

Calculate the value at t=0.1:

For t₀=0 and y₀=1:

f(0,1)= \cdot \sin(0) = 0

y_1 = y_0 + h \cdot f(0, 1) = 1 + 0.1 \cdot 0 = 1

Thus, y(0.1) \approx 1.

Calculate the value at t=0.2:

For t₁=0.1 and y₁=1:

f(0.1, 1) = 1 \cdot \sin(0.1) \approx 0.0998

y_2 = y_1 + h \cdot f(0.1, 1) = 1 + 0.1 \cdot 0.0998 \approx 1 + 0.00998 = 1.00998

Thus, y(0.2) \approx 1.00998.

P3. ODE: \frac{dy}{dt} = t - y

Initial Condition: y(1)=2

Step Size (h): 0.5

Solution:

Calculate the value at t=1.5:

For t₀=1 and y₀=2:

f(1,2)=1−2=−1

y_1 = y_0 + h \cdot f(1, 2) = 2 + 0.5 \cdot (-1) = 2 - 0.5 = 1.5

Thus, y(1.5) \approx 1.5.

Calculate the value at t=2.0:

For t₁=1.5 and y₁=1.5:

f(1.5, 1.5) = 1.5 - 1.5 = 0

y_2 = y_1 + h \cdot f(1.5, 1.5) = 1.5 + 0.5 \cdot 0 = 1.5

Thus,y(2.0) \approx 1.5.

P4. ODE: \frac{dy}{dt} = e^t - y

Initial Condition: y(0)=0

Step Size (h): 0.2

Solution:

Calculate the value at t=0.2:

For t₀=0 and y₀=0:

f(0, 0) = e⁰ - 0 = 1

y_1 = y_0 + h \cdot f(0, 0) = 0 + 0.2 \cdot 1 = 0.2

Thus, y(0.2) \approx 0.2.

Calculate the value at t=0.4:

For t₁=0.2 and y₁=0.2:

f(0.2, 0.2) = e^{0.2} - 0.2 \approx 1.2214 - 0.2 = 1.0214

y_2 = y_1 + h \cdot f(0.2, 0.2) = 0.2 + 0.2 \cdot 1.0214 = 0.2 + 0.20428 = 0.40428

Thus, y(0.4) \approx 0.40428.

P5. ODE: \frac{dy}{dt} = t^2 - y^2

Initial Condition: y(0)=1

Step Size (h): 0.1

Solution:

Calculate the value at t=0.1:

For t₀=0 and y₀=1:

f(0, 1) = 0^2 - 1^2 = -1

y_1 = y_0 + h \cdot f(0, 1) = 1 + 0.1 \cdot (-1) = 1 - 0.1 = 0.9

Thus, y(0.1) \approx 0.9.

Calculate the value at t=0.2:

For t₁=0.1 and y₁=0.9:

f(0.1, 0.9) = 0.1² - 0.9² = 0.01 - 0.81 = -0.8

y_2 = y_1 + h \cdot f(0.1, 0.9) = 0.9 + 0.1 \cdot (-0.8) = 0.9 - 0.08 = 0.82

Thus, y(0.2) \approx 0.82.

P6. ODE: \frac{dy}{dt} = 2t - y

Initial Condition: y(0)=1

Step Size (h): 0.1

Solution:

Calculate the value at t=0.1:

For t0=0 and y₀=1:

f(0, 1) = 2 \cdot 0 - 1 = -1

y_1 = y_0 + h \cdot f(0, 1) = 1 + 0.1 \cdot (-1) = 1 - 0.1 = 0.9

Thus, y(0.1) \approx 0.9.

Calculate the value at t=0.2:

For t₁=0.1 and y₁=0.9:

f(0.1, 0.9) = 2 \cdot 0.1 - 0.9 = 0.2 - 0.9 = -0.7 y_2 = y_1 + h \cdot f(0.1, 0.9) = 0.9 + 0.1 \cdot (-0.7) = 0.9 - 0.07 = 0.83 Thus, y(0.2) \approx 0.83.

P7. ODE: \frac{dy}{dt} = -3y + 4t

Initial Condition: y(0)=2

Step Size (h): 0.2

Solution:

Calculate the value at t=0.2:

For t₀=0 and y₀=2:

f(0, 2) = -3 \cdot 2 + 4 \cdot 0 = -6

y_1 = y_0 + h \cdot f(0, 2) = 2 + 0.2 \cdot (-6) = 2 - 1.2 = 0.8

Thus, y(0.2) \approx 0.8.

Calculate the value at t=0.4:

For t₁=0.2 and y₁=0.8:

f(0.2, 0.8) = -3 \cdot 0.8 + 4 \cdot 0.2 = -2.4 + 0.8 = -1.6

y_2 = y_1 + h \cdot f(0.2, 0.8) = 0.8 + 0.2 \cdot (-1.6) = 0.8 - 0.32 = 0.48

Thus, y(0.4) \approx 0.48.

P8. ODE: \frac{dy}{dt} = t^2 + y

Initial Condition: y(1)=0

Step Size (h): 0.5

Solution:

Calculate the value at t=1.5:

For t₀=1 and y₀=0:

f(1, 0) = 1² + 0 = 1

y_1 = y_0 + h \cdot f(1, 0) = 0 + 0.5 \cdot 1 = 0.5

Thus, y(1.5) \approx 0.5.

Calculate the value at t=2.0:

For t₁=1.5 and y₁=0.5:

f(1.5, 0.5) = 1.5² + 0.5 = 2.25 + 0.5 = 2.75

y_2 = y_1 + h \cdot f(1.5, 0.5) = 0.5 + 0.5 \cdot 2.75 = 0.5 + 1.375 = 1.875

Thus, y(2.0) \approx 1.875.

P9. ODE: \frac{dy}{dt} = \cos(t) - y

Initial Condition: y(0)=0

Step Size (h): 0.1

Solution:

Calculate the value at t=0.1:

For t₀=0 and y₀=0:

f(0,0)=cos⁡(0)−0=1

y_1 = y_0 + h \cdot f(0, 0) = 0 + 0.1 \cdot 1 = 0.1

Thus, y(0.1) \approx 0.1.

Calculate the value at t=0.2:

For t₁=0.1 and y₁=0.1:

f(0.1, 0.1) = \cos(0.1) - 0.1 \approx 0.995 - 0.1 = 0.895

y_2 = y_1 + h \cdot f(0.1, 0.1) = 0.1 + 0.1 \cdot 0.895 = 0.1 + 0.0895 = 0.1895

Thus, y(0.2) \approx 0.1895.

P10. ODE: \frac{dy}{dt} = -y^2 + t\frac{dy}{dt} = -y^2 + t

Initial Condition: y(0)=2

Step Size (h): 0.3

Solution:

Calculate the value at t=0.3t = 0.3t=0.3:

For t₀=0 and y₀=2:

f(0, 2) = -2² + 0 = -4

y_1 = y_0 + h \cdot f(0, 2) = 2 + 0.3 \cdot (-4) = 2 - 1.2 = 0.8

Thus, y(0.3) \approx 0.8.

Calculate the value at t=0.6:

For t₁=0.3 and y₁=0.8:

f(0.3, 0.8) = -0.8² + 0.3 = -0.64 + 0.3 = -0.34

Read More,

• Euler’s Formula
• Number Theory
• Prime Numbers
• Euler’s Theorem

Eulers Method Practice Questions

Q1. Use the Euler method with a step size of h=0.1 to approximate the solution of the initial value problem \frac{dy}{dt} = y, y(0)=1 at t=0.2.

Q2. Use the Euler method to approximate the solution of \frac{d^2y}{dt^2} = -y with initial conditions y(0)=1 and \frac{dy}{dt}(0) = 0 at t=0.1, using h=0.01.

Q3. Approximate the solution of \frac{dy}{dt} = y^2, y(0)=1 at t=0.1 using the Euler method with h=0.05.

Q4. Use the Euler method to solve the system:

\frac{dx}{dt} = x - y

\frac{dy}{dt} = x + y

with initial conditions x(0)=1, y(0)=0, and h=0.1 at t=0.2.

Q5. Approximate the solution of \frac{dy}{dt} = t - y, y(0)=0 at t=0.2 using h=0.1.

Q6. ODE: \frac{dy}{dt} = 3t - y

Initial Condition: y(0)=1

Step Size (h): 0.1

Q7. ODE: \frac{dy}{dt} = \cos(t) - y

Initial Condition: y(0)=1

Step Size (h): 0.1

Q8. ODE: \frac{dy}{dt} = y \cdot \sin(t)

Initial Condition: y(0)=2

Step Size (h): 0.3

Q9. ODE: \frac{dy}{dt} = y \cdot \sin(t)

Initial Condition: y(0)=0

Step Size (h): 0.01

Q10. ODE: \frac{dy}{dt} = \cos(t) - y

Initial Condition: y(0)=2

Step Size (h): 0.3