# Homogeneous initial value problem

• mtxop2
In summary, the equation given is a homogeneous equation with initial conditions y(0) = 3 and y'(0) = 1. The solution involves finding the roots of the equation and using the formula yh = e^ax(c1cosbx + c2sinbx). The attempt at a solution involves differentiating this formula using the product rule to obtain a more simplified equation.
mtxop2

## Homework Statement

4y" + 4y' + 5y = 0
y(0) = 3
y'(0) = 1

## Homework Equations

yh = e^ax(c1cosbx + c2sinbx)

## The Attempt at a Solution

For the roots I got -1/2 + i and -1/2 - i so my a = -1/2 and b = 1

then I have to differentiate yh = e^(-1/2x)[c1cosx + c2sinx]

this is where I get this overly complicated equation and I was wondering if I could do integration by parts instead.

where I could get something like

yh = -1/2e^(-1/2x)[c1cosx + c2sinx] + [-c1sinx + c2cosx]e^(-1/2x)

Last edited:
mtxop2 said:

## Homework Statement

4y" + 4y' + 5y = 0
y(0) = 3
y'(0) = 1

## Homework Equations

yh = e^ax(c1cosbx + c2sinbx)

## The Attempt at a Solution

For the roots I got -1/2 + i and -1/2 - i so my a = -1/2 and b = 1

then I have to differentiate yh = e^(-1/2x)[c1cosx + c2sinx]

this is where I get this overly complicated equation and I was wondering if I could do integration by parts instead.
?
You can't use integration by parts if you need to differentiate a function.
mtxop2 said:
where I could get something like

yh = -1/2e^(-1/2x)[c1cosx + c2sinx] + [-c1sinx + c2cosx]e^(-1/2x)

Mark44 said:
?
You can't use integration by parts if you need to differentiate a function.

Damn I'm so sorry I misled you, I've been working on these all day and well you know how that goes. What I meant to say was product rule, where f(x) = e^(-1/2x) and g(x) = [c1cosx + c2sinx]

Yes, use product rule for y= e-x/2(c1cosx+c2sinx) to get y'.

ehild

## 1. What is a homogeneous initial value problem?

A homogeneous initial value problem is a type of differential equation that involves finding a function that satisfies both the equation and a set of initial conditions. The term "homogeneous" refers to the fact that the equation is linear and does not contain any constant terms.

## 2. How do you solve a homogeneous initial value problem?

To solve a homogeneous initial value problem, you can use the method of separation of variables or the method of undetermined coefficients. Both methods involve finding a general solution to the equation and then using the initial conditions to determine the specific solution that satisfies the given conditions.

## 3. What are the applications of homogeneous initial value problems?

Homogeneous initial value problems are commonly used in various fields of science, such as physics, engineering, and economics, to model and analyze various phenomena like population growth, heat transfer, and electrical circuits.

## 4. What is the difference between a homogeneous and non-homogeneous initial value problem?

The main difference between a homogeneous and non-homogeneous initial value problem is that the latter contains a constant term, while the former does not. This term can be a function of the independent variable or a constant, and it makes the equation more complex to solve.

## 5. Can a homogeneous initial value problem have multiple solutions?

Yes, a homogeneous initial value problem can have multiple solutions. This is because the general solution to the equation usually contains arbitrary constants, and different choices for these constants can result in different solutions that satisfy the given initial conditions.

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