Method of Undetermined Coefficients

[_N3WTON_]

Homework Statement

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

Homework Equations

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?

 

[SteamKing]

Homework Statement

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

Homework Equations

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.

 

[_N3WTON_]

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