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I am investigating insofar to the general solution of a basic problem. From what I've researched, I need to use a special method to solve the following differential equation

$$y''-y'-2y = e^{3x}$$

We have a complementary solution of

$$y_c = c_1e^{2x} + c_2e^{-x}$$

with $a=2, b=1$ by quadratic (or factoring). We presume that $y_1=p(x)e^{3x}$ is a particular solution and differentiate, yielding

$$y_1'=p'(x)e^{3x} + 3p(x)e^{3x}$$, $$y_1''=p''(x)e^{3x}+3e^{3x}p'(x)+3p'(x)e^{3x}+9p(x)e^{3x}$$

Simplifying $y_1''$ we have

$$y_1'' = p''(x) + 6p'(x) + 9p(x) = 1$$

dividing by $e^{3x}$ is obvious.

But now what do I do? I feel I have left out integration. Is this the correct usage of Variation of Parameters, for Differential Equations?

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  • $\begingroup$ What's wrong with trying $y_p=Ce^{3x}$? $\endgroup$
    – lulu
    Jul 17, 2017 at 22:21
  • $\begingroup$ I have some arithmetic errors. It's no wonder I did not realize that sooner. No wonder! Again, I apologize for posting posthumously. $\endgroup$
    – cryptomath
    Jul 17, 2017 at 22:34
  • $\begingroup$ Just to say: "posthumously" refers to things done after death...so I truly hope you are not posting anything posthumously! $\endgroup$
    – lulu
    Jul 18, 2017 at 0:37

2 Answers 2

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Instead of $p(x)$ as a function. Make it a constant.

$y_1 = p e^{3x}\\ y_1' = 3p e^{3x} \\ y_1'' = 9p e^{3x}$

now plug these into the original diff eq.

$y_1''-y_1' - 2y_1 = 4pe^{3x} = e^{3x}$

and solve for $p$

$p = \frac 14$

$y = c_1 e^{-x} + c_2 e^{2x} + \frac 14 e^{3x}$

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  • $\begingroup$ Sorry, I posted posthumously. Thanks for your answer. Very straightforward. $\endgroup$
    – cryptomath
    Jul 17, 2017 at 22:32
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To eliminate the exponential term, and get an easier equation, begin by putting $$y=ze^{3x} $$ $$y'=(z'+3z)e^{3x} $$ $$y''=(z''+6z'+9z)e^{3x} $$

the equation becomes

$$z''+5z'+4z=1$$

the characteristic equation is $$r^2+5r+4=0$$ the roots are $$r_1=-4 \;,\;r_2=-1$$ a particular solution is $$z_p=\frac {1}{4} $$ the general solution is $$y=(\lambda e^{-4x}+\mu e^{-x}+\frac {1}{4})e^{3x}$$ $$=\lambda e^{-x}+\mu e^{2x}+\frac {e^{3x}}{4} $$

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  • $\begingroup$ Whenever you would accept, I would like to have a chat with you. $\endgroup$ Jul 18, 2017 at 5:12

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