I need guidance to complete this problem properly. I have

$$y'' - 7y' = -3$$

I am asked to find the general solution. To do that, I first start off by finding the characteristic polynomial

$$y_c = c_1e^{7x} + c_2$$

where $r=0, r=7$. A particular solution could perhaps resemble

$$y_p = u_1e^{7x} + u_2$$


$$u_1'e^{7x} + u_2' = 0$$ then $$7u_1'e^{7x} = -3$$

Adding the two equations yeilds:

$$u_1'e^{7x}=u_2' => u_1'=-u_2'e^{-7x} = -3$$

Then I integrated to find $u_1, u_2$

$$u_1 = -\int{3dx} = 3x$$ $$u_2 = \int{e^{7x}} = \frac{1}{7}e^{7x}$$

So my particular solution, $y_p$ is

$$y_p = u_1e^{7x}+u_2 = -3xe^{7x} + \frac{1}{7}e^{7x}$$

I found the general solution as

$$y = c_1e^{7x} + c_2 - 3xe^{7x} + \frac{1}{7}e^{7x}$$

but that is incorrect. What is left undone? What have I done wrong? Is this the right methodology/technique to be using?

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    $\begingroup$ In the second equation you should not have a $u_2'$ the derivative of a constant is 0 $\endgroup$ – Teh Rod Jul 18 '17 at 18:15
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    $\begingroup$ I suggest that the particular solution is a polynomial. $\endgroup$ – Doug M Jul 18 '17 at 18:34
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    $\begingroup$ Why is your particular solution so complicated? Surely $y_p=Cx$ works. $\endgroup$ – lulu Jul 18 '17 at 18:38
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    $\begingroup$ Note: as with your prior question you appear to overcomplicate the search for particular solutions. I'd practice that. $\endgroup$ – lulu Jul 18 '17 at 18:39
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    $\begingroup$ No...The differential operator on the left is $D[y]=y''-7y$. Thus $D[Cx]=0-7\times C=-7C$. Thus you want $-7C=-3$ or $C=\frac 37$. You already have the homogeneous solution (well, I didn't check but it is straight forward). $\endgroup$ – lulu Jul 18 '17 at 18:49

The best way to solve this one would be to set $v=y'\implies v'=y''$ the equation becomes $v'-7v=-3$ which is separable.


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