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.


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..


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

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


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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.