Method of Undetermined Coefficients

[_N3WTON_]

Homework Statement

Find the general solution by finding the homogeneous solution and a particular solution.
[itex] y'' + 4y' = x [/itex]

Homework Equations

The Attempt at a Solution

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

Homework Equations

The Attempt at a Solution

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