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Suppose I have a equation

$$y''+\alpha y'+\beta y = f(x)$$

Characteristic equation should be

$$\lambda^2 + \alpha\lambda + \beta = 0$$ $$\lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ $$where \ a=1, b=\alpha, c=\beta$$

If my general solution is given to be $$y_h = e^{3x}(C_1 \sin 4x + C_2 \cos 4x)$$

Does that mean that $$3= \frac{-b}{2a}$$ $$4i = \frac{\sqrt{b^2 -4ac}}{2a}$$

Please explain, thanks.

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This kind of solution is obtained when the auxiliary equation obtained has complex roots. Say, $p+iq$ and $p-iq$. Now, the general solution of the homogenous DE is $Ae^{(p+iq)x}+Be^{(p-iq)x}$. Now, since $e^{iqx}=cos qx+isinqx$, we have the solution as $e^{px}(C_1cosqx+C_2sinqx)$ where, $C_1=A+B $ and $C_2=i(A-B)$. Now, let's go back to the equation: $a\lambda^2+b\lambda+c=0$. You have $p=-b/a$ and $p^2+q^2=c/a$. Now, can you figure it out from here..

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Say a Characteristic equation $a\lambda^2+b\lambda+c=0$ has solutions $p\pm iq$. The Imaginary component must come from the discriminant being negative i.e. $b^2-4ac<0$.

So by equating both sides (the solution itself and the generic solution of a Quadratic equation),

$$ p\pm iq = \frac{-b\pm\sqrt{b^2-4ac}}{2a} $$

you can assert that

$$ \text{Re}\left(p\pm iq\right) = -\frac{b}{2a} $$

and

$$ \text{Im}\left(p\pm iq\right) =\pm\frac{\sqrt{b^2-4ac}}{{2a}} $$

provided that $b^2-4ac<0$.

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