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