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I'm struggling with this differential equation: $ y' -3y = x \cdot e^x $
Step one: Solve the homogenous equation $ y' -3y = 0 $ $$ y_0 = e^{3x} \cdot C $$

The approach for the particular solution $y_p$ should be $(c_1x+c_0)\cdot Ce^x$

This leads me to the following term: $$ y_p = Ce^x(ax+b) $$ $$ y'_p= Ce^x\cdot(ax+b)+Ce^x\cdot a $$ $$ Ce^x\cdot(ax+b)+Ce^x\cdot a - 3Ce^x(ax+b) = x\cdot e^x $$ $$ C\cdot(ax+b)+C\cdot a - 3C(ax+b) = x $$

The C is a real problem here. After taking a look in my textbook I saw that the approach for $y_p$ is: $$y_p = (ax + b)\cdot e^x $$ $$y'_p = e^x \cdot (ax +b)+e^x \cdot a $$ Plugging this into $ y' - 3y = x\cdot e^x $ and simplify gives:
$$ x(-2a)+a-2b=x \Rightarrow a=-\frac{1}{2}; b=-\frac{1}{4}$$ And finally $$ y= e^{3x}\cdot C - \frac{1}{4}(e^x+2xe^x) $$

When I omit the C I'm able to solve this but I don't know why this is correct as the approach for $g(x)$ would be $C\cdot e^{bx}$

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    $\begingroup$ The particular solution is supposed to be any solution to the whole equation. $C$ is extraneous as you can absorb it into $a$ and $b$. You can see that by looking at your solution where it always multiplies one or the other. If you are going to use $C$ you should use a different variable than the one you used for the general solution. That is a good way to get confused. $\endgroup$ – Ross Millikan Sep 19 '18 at 0:16
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Just note that as $c_0 , c_1 , C$ vary all over $\mathbb{R}$ you have that $a=c_1C$ and $b=c_0C$ vary all over $\mathbb{R}$ AND $(c_1x+c_0)Ce^x = (ax+b)e^x$.

So considering $3$ variables is useless (they are too many) since you only want a constant for $xe^x$ and a constant for $e^x$.

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You have complicated things but your idea gives the same answer as the textbook

$$C\cdot(ax+b)+C\cdot a - 3C(ax+b) = x$$ $$-2Cax+(Ca-2Cb)=x$$ $$\implies aC=-1/2,$$ $$ \implies 2Cb=Ca \implies Cb=-1/4 $$ $$y_p = Ce^x(ax+b) \implies y_p = e^x(-\frac12x-\frac 14)$$ $$y_p = -\frac 12e^x(x+\frac 12)$$

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