# Max Height Homogenous Differential Equation

• shards5
In summary: Then find the maximum value of y(x) using the derivative of this general solution.In summary, the solution to this initial value problem is y(x) = c0cosx + c1sinx + (1/3)sin2x, with c0 and c1 being constants determined by the initial conditions y(0) = 0 and y'(0) = 0. The maximum value of y is found by taking the derivative of the general solution and setting it equal to zero, giving a maximum value of y = 1/3 when x = pi/4.
shards5

## Homework Statement

Let y be the the solution of the initial value problem. The maximum value of y is?
y" + y = -sin(2x)
y(0) = 0
y'(0) = 0

N/A

## The Attempt at a Solution

Factor out original equation to find general equation.
r2 + 1 = 0
r = 0 + i; 0- i
Thus the general solution is.
y(x) = c0cos(x) + c1sin(x)
y'(x) = -c0sin(x) + c1cos(x)
Using the initial conditions I found that.
c0 = 0
c1 = 1
Now to find A*cos(2x) + Bsin(2x)
yp = A*cos(2x) + B*sin(2x)
y'p = -2A*sin(2x) + 2B*cos(2x)
y"p = -4A*cos(2x) - 4B*sin(2x)
Plugging into the original equation we get.
-4A*cos(2x) - 4B*sin(2x) + A*cos(2x) + B*sin(2x) = -sin(2x)
which gives us with -3Bsin(2x) = -sin(2x) B = 1/3 and A = 0
Therefore the equation is.
y(x) = 1/3*sin(2x)
To find the maximum value I do the derivative of the above equation and set it equal to zero.
y'(x) = 2/3*cos(2x) = 0 which gives x to be pi/4
Using x = pi/4 in the unique equation.
y(pi/4) = 1/3*sin(2*pi/4)
I get y = 0.333333333333333 which is wrong, of course.
So what am I doing wrong?

You're working only with the particular solution, not the general solution. Your general solution is y(x) = sin(x) + (1/3)sin(2x). Find the maximum value of that function.

I made a mistake c1 is supposed to also equal zero so my particular solution would only have 1/3sin(2x).

The initial conditions are for the general solution y(x), not just for the complementary solution. Start with a general solution of y(x) = c0cosx + c1sinx + (1/3)sin2x and calculate the constants c0 and c1 using your two initial conditions.

## 1. What is a max height homogenous differential equation?

A max height homogenous differential equation is a type of differential equation that models the behavior of a system by tracking the rate of change of a variable with respect to time. This type of equation is "homogenous" because all of the terms in the equation have the same degree, meaning they all have the same power.

## 2. How is a max height homogenous differential equation different from a regular differential equation?

A regular differential equation can have terms with different degrees, while a max height homogenous differential equation has all terms with the same degree. This makes it easier to solve because it can be rewritten in a simpler form by substituting a new variable.

## 3. What is the importance of solving a max height homogenous differential equation?

Solving a max height homogenous differential equation allows us to understand the behavior of a system over time. This is important in various fields of science, such as physics, chemistry, and biology, as it can help us predict and control the behavior of systems in the real world.

## 4. How do you solve a max height homogenous differential equation?

To solve a max height homogenous differential equation, you can use the substitution method, where a new variable is introduced to simplify the equation. This new variable is usually chosen based on the degree of the original equation. Once the equation is in a simpler form, it can be solved using techniques such as separation of variables or integrating factors.

## 5. What are some real-world applications of max height homogenous differential equations?

Max height homogenous differential equations have many real-world applications, such as modeling population growth, chemical reactions, and radioactive decay. They are also used in engineering to study the behavior of systems like electrical circuits and mechanical systems. In physics, they are used to describe the motion of objects under the influence of forces.

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