# Y''' + 8y = 2x - 5 + 8 e^(-2x); y(0) = -5, y'(0) = 3, y''(0) = -4

• s3a
In summary, the student is trying to solve an equation, but is making the same mistake over and over.
s3a

## Homework Statement

The problem:
y''' + 8y = 2x - 5 + 8 e^(-2x); y(0) = -5, y'(0) = 3, y''(0) = -4

y = -23/12 e^(-2x) + e^x (-59/24 cos(sqrt(3) x) + 17 sqrt(3)/72 sin(sqrt(3) x)) + 1/4 x - 5/8 + 2/3 x e^(-2x)

## Homework Equations

Method of undetermined coefficients.

## The Attempt at a Solution

My work is attached as the TheProblemAndMyWork.pdf file. The work is typed (not handwritten), so it looks nice.

What I'm having trouble with is finding the values for c_1, c_2 and c_3. I entered the system of equations from my work into Wolfram Alpha, and the answer is wrong! I re-did this problem so many times, and I keep ending up in the same situation. I suppose, I keep making the exact same mistake.

Could someone please tell me what I am doing wrong?

Any help would be GREATLY appreciated!

#### Attachments

• TheProblemAndMywork.pdf
51.4 KB · Views: 1,853
The constant term you get when you calculated f''(0) should be -8/3, not 8.

s3a said:

## Homework Statement

The problem:
y''' + 8y = 2x - 5 + 8 e^(-2x); y(0) = -5, y'(0) = 3, y''(0) = -4

y = -23/12 e^(-2x) + e^x (-59/24 cos(sqrt(3) x) + 17 sqrt(3)/72 sin(sqrt(3) x)) + 1/4 x - 5/8 + 2/3 x e^(-2x)

## Homework Equations

Method of undetermined coefficients.

## The Attempt at a Solution

My work is attached as the TheProblemAndMyWork.pdf file. The work is typed (not handwritten), so it looks nice.

What I'm having trouble with is finding the values for c_1, c_2 and c_3. I entered the system of equations from my work into Wolfram Alpha, and the answer is wrong! I re-did this problem so many times, and I keep ending up in the same situation. I suppose, I keep making the exact same mistake.

Could someone please tell me what I am doing wrong?

Any help would be GREATLY appreciated!

It is a quite a bit easier if you first get rid of the x and constant terms on the right. You can do this by looking at ##z(x) = y(x) + ax + b## for some constants ##a, b##, then adjusting ##a## and ##b## until you get ##z''' + 8z = 8 e^{ -2x}##. Then the "undetermined coefficient" equations simplify down quite a lot. Of course you need to determine the initial conditions on z(t) and use them instead.

Thanks guys, I get it now!

Could someone explain to me how the exercise found in pdf is done, please thank you.

That’s not how it works here. You need to show an attempt, and we can help you identify and correct your mistakes and misconceptions.

There should be plenty of examples you can see in your textbook.

Mark44

## 1. What is the equation for "Y''' + 8y = 2x - 5 + 8 e^(-2x)"?

The equation is a third-order linear differential equation, also known as a third-order ordinary differential equation (ODE), and is expressed as:

Y''' + 8y = 2x - 5 + 8 e^(-2x)

## 2. What are the initial conditions for this equation?

The initial conditions for this equation are given as y(0) = -5, y'(0) = 3, and y''(0) = -4. These are known as the initial value conditions, which specify the values of the dependent variable (y) and its derivatives at a specific point (x = 0) in the domain.

## 3. What is the meaning of "e^(-2x)" in the equation?

"e^(-2x)" is the exponential function with a base of e (Euler's number) raised to the power of -2x. It represents the rate of change of a quantity over time and is often used in natural growth or decay problems in mathematics and science.

## 4. How do you solve this differential equation?

First, rearrange the equation to isolate the dependent variable (y), then integrate the equation three times to obtain the general solution. Next, use the initial conditions to find the particular solution that satisfies the given conditions. Finally, substitute the values of the particular solution to obtain the final solution that solves the differential equation.

## 5. What is the practical application of this equation?

This equation can be used to model real-life situations where the rate of change of a quantity (y) is affected by an external stimulus (2x - 5) and an exponential decay term (e^(-2x)). Some examples include population growth or decay, radioactive decay, and chemical reactions.

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