# 2nd order differential equation

• NEGATIVE_40
In summary, the conversation is about solving a linear non-homogeneous differential equation with the given equation y''+16y=3.36. The approach used was to first find the homogeneous solution using the characteristic equation, and then find the particular solution by substituting a trial function of the form yp=A and solving for A. The rationale for choosing this particular trial function is to avoid having a solution that is a multiple of the homogeneous solution.
NEGATIVE_40

## Homework Statement

y''+16y=3.36

This is actually part of a spring question I'm attempting at the moment, and I'm having a mental blank on how to deal with the 3.36.

n/a

## The Attempt at a Solution

I've found the characteristic equation and solution based from that;
$$c_1 cos(4x) + c_2 sin(4x)$$
$$c_1 cos(4x) + c_2 sin(4x) + 0.21$$

Last edited:
Remember: The general solution to any linear non-homogeneous differential eq. is equal to the soln. to the homogeneous sq. plus the particular soln. to the non-homogeneous eq.
What is your characteristic eq. here?

You found the homogeneous solution. You need to find the particular solution. Try substituting a trial solution of the form yp=A and solving for A.

vela said:
You found the homogeneous solution. You need to find the particular solution. Try substituting a trial solution of the form yp=A and solving for A.

[tex] y_p = A [tex]
so doing that I get A = 0.21, which gives me the correct solution.

thanks.

However do you know the rationale for choosing trail function as A instead of A x

yep. so it isn't a multiple of the homogenous solution. I did a course of DE's a while ago, I just forgot what to do in this case.

icystrike said:
However do you know the rationale for choosing trail function as A instead of A x
If you had tried Ax, then (Ax)''= 0 so the equation would have been 16Ax= 3.36. The left side has an "x" and the right side doesn't. May I ask why you are doing this problem? I would imagine it is because you are taking a course in differential equations but I cannot imagine you having a homework problem like this if you are not in a chapter of your textbook that has a detailed description of "Undetermined Coefficients" as this method is called.

## 1. What is a 2nd order differential equation?

A 2nd order differential equation is a mathematical equation that involves a function and its two derivatives. It can be written in the form: y'' + f(x)y' + g(x)y = h(x), where y'' represents the second derivative of y, f(x) and g(x) are functions of x, and h(x) is a function of x.

## 2. What is the difference between a 1st order and a 2nd order differential equation?

A 1st order differential equation involves only the first derivative of the function, while a 2nd order differential equation involves both the first and second derivatives. This means that a 2nd order differential equation is more complex and requires more information to solve.

## 3. What are some real-life applications of 2nd order differential equations?

2nd order differential equations are used in many fields of science and engineering. Some examples include modeling the motion of a pendulum, predicting the growth of a population, and analyzing the behavior of electrical circuits.

## 4. How do you solve a 2nd order differential equation?

Solving a 2nd order differential equation involves finding a function that satisfies the given equation. This can be done analytically using various methods such as separation of variables, substitution, or integrating factors. It can also be solved numerically using computer software or techniques such as Euler's method.

## 5. What is the significance of the initial conditions in solving a 2nd order differential equation?

The initial conditions, also known as boundary conditions, are values of the function and its derivatives at a specific point. These conditions are necessary to find the unique solution to a 2nd order differential equation. Without them, the solution would be a general expression with an infinite number of possible solutions.

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