# What is the general solution to a differential equation?

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

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

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

and

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

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