Answer to Question #215964 in Differential Equations for M.shazil

Question

Discuss in detail all cases of the roots of a second order linear differential equation with constant

coefficients.

Answer 1

Given the second order linear homogeneous equations with constant coefficients

$$ay''+by'+c=0, a\not=0$$

Let the characteristic equation of the differential equation be

$$ar^2+br+c=0$$

Case 1 Two distinct real roots

When $b^2-4ac>0,$ the characteristic polynomial has two distinct real roots $r_1, r_2.$ They give two distinct solutions $y_1=e^{r_1t}$ and $y_2=e^{r_2t}.$

Therefore, a general solution of the second order linear homogeneous equations with constant coefficients is

$$y(t)=c_1e^{r_1t}+c_2e^{r_2t}$$

Case 2 Two complex conjugate roots

When $b^2-4ac<0,$ the characteristic polynomial has two complex roots, which are conjugates, $r_1=\lambda+\mu i, r_2=\lambda-\mu i$ ($\lambda,\mu$ are real numbers, $\mu>0$ ).

As before they give two linearly independent solutions $y_1=e^{r_1t}$ and $y_2=e^{r_2t}.$ Consequently the linear combination $y(t)=c_1e^{r_1t}+c_2e^{r_2t}$ will be a general solution. 

Using the Euler's Formula we have

$$y(t)=c_3e^{\lambda t}\cos (\mu t)+c_4e^{\lambda t}\sin (\mu t)$$

Case 3 One repeated real root

When $b^2-4ac=0,$ the characteristic polynomial has a single repeated real root $r=-\dfrac{b}{2a}.$ The general solution in the case of a repeated real root $r$ is

$$y(t)=c_1e^{rt}+c_2te^{rt}.$$