# Method of Undetermined Coefficients

• _N3WTON_
In summary: So my solution is correct. In summary, the general solution of the differential equation y'' + 4y' = x is Y(x) = c_1 + c_2e^{-4x} + \frac{x^2}{8} - \frac{x}{16}.
_N3WTON_

## Homework Statement

Find the general solution by finding the homogeneous solution and a particular solution.
$y'' + 4y' = x$

## The Attempt at a Solution

First, I found the corresponding solution to the homogeneous differential equation:
$y'' + 4y' = 0$
$r^{2} + 4r = 0$
$r_1 = 0$
$r_2 = -4$
$Y_{h}(x) = c_1 + c_{2}e^{-4x}$
Now, I applied the method of undetermined coefficients:
$Y_{p}(x) = Ax^{2} + Bx + C$
$Y_{p}'(x) = 2Ax + B$
$Y_{p}''(x) = 2A$
I substituted these values into the original differential equation and found:
$2A + 4(2Ax + B) = x$
$2A + 8Ax + 4B = x$
So, I obtained the following system of equations:
$8A = 1$
$2A + 4B = 0$
Therefore, I found:
$A = \frac{1}{8} \hspace{2 mm} {and} \hspace{2 mm} B = -\frac{1}{16}$
So a particular solution of the differential equation is:
$Y_{p}(x) = \frac{x^2}{8} - \frac{x}{16}$
And the general solution of the differential equation is:
$Y(x) = c_1 + c_{2}e^{-4x} + \frac{x^2}{8} - \frac{x}{16}$
I believe this is correct; however, the solution given in the back of my textbook is:
$Y(x) = c_1 + c_{2}e^{-4x} + \frac{x^2}{8} - \frac{1}{16}$
Is this a mistake? Or did I make a mistake somewhere?

_N3WTON_ said:

## Homework Statement

Find the general solution by finding the homogeneous solution and a particular solution.
$y'' + 4y' = x$

## The Attempt at a Solution

First, I found the corresponding solution to the homogeneous differential equation:
$y'' + 4y' = 0$
$r^{2} + 4r = 0$
$r_1 = 0$
$r_2 = -4$
$Y_{h}(x) = c_1 + c_{2}e^{-4x}$
Now, I applied the method of undetermined coefficients:
$Y_{p}(x) = Ax^{2} + Bx + C$
$Y_{p}'(x) = 2Ax + B$
$Y_{p}''(x) = 2A$
I substituted these values into the original differential equation and found:
$2A + 4(2Ax + B) = x$
$2A + 8Ax + 4B = x$
So, I obtained the following system of equations:
$8A = 1$
$2A + 4B = 0$
Therefore, I found:
$A = \frac{1}{8} \hspace{2 mm} {and} \hspace{2 mm} B = -\frac{1}{16}$
So a particular solution of the differential equation is:
$Y_{p}(x) = \frac{x^2}{8} - \frac{x}{16}$
And the general solution of the differential equation is:
$Y(x) = c_1 + c_{2}e^{-4x} + \frac{x^2}{8} - \frac{x}{16}$
I believe this is correct; however, the solution given in the back of my textbook is:
$Y(x) = c_1 + c_{2}e^{-4x} + \frac{x^2}{8} - \frac{1}{16}$
Is this a mistake? Or did I make a mistake somewhere?

It should be fairly easy for you to check which solution satisfies the original ODE.

SteamKing said:
It should be fairly easy for you to check which solution satisfies the original ODE.
sorry, I didn't even think of that...I did this and found that the book's solution is incorrect

## What is the Method of Undetermined Coefficients?

The Method of Undetermined Coefficients is a mathematical technique used to solve non-homogeneous linear differential equations. It involves finding a particular solution by assuming the form of the solution and solving for the undetermined coefficients.

## When is the Method of Undetermined Coefficients used?

The Method of Undetermined Coefficients is used when solving non-homogeneous linear differential equations with constant coefficients. It is most commonly used when the non-homogeneous term has a simple form, such as a polynomial, exponential, or trigonometric function.

## How does the Method of Undetermined Coefficients work?

The Method of Undetermined Coefficients works by assuming a particular form for the solution and then substituting it into the original differential equation. The undetermined coefficients are then solved for by equating coefficients of like terms on both sides of the equation.

## What are the limitations of the Method of Undetermined Coefficients?

The Method of Undetermined Coefficients is limited to solving non-homogeneous linear differential equations with constant coefficients. It also only works when the non-homogeneous term has a simple form. In some cases, the assumed form of the solution may need to be modified to include additional terms.

## Are there any alternative methods to solve non-homogeneous linear differential equations?

Yes, there are alternative methods such as the Method of Variation of Parameters and Laplace Transform. These methods may be more useful in cases where the non-homogeneous term is more complex or when the differential equation is in a non-standard form.

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