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## 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?