Concept:

For a second-order differential function, the characteristic equation will have two roots D1 and D2. The roots can have three possible forms i.e.

• Real, distinct roots, D1 ≠ D2
• Complex roots, D1, D2 = a ± ib
• Double roots D1 = D2 = 0

The solution for the second-order differential equation with equal roots of the characteristic equation is given by:

\(y = \left( {{c_1} + {c_2}x} \right){e^{{D_1x}}}\)

If two roots are real and distinct (i.e. D1 ≠ D2), the general solution will be:

\(y = {C_1}{e^{{D_1}x}} + {C_2}{e^{{D_2}x}}\)

If the roots of the characteristic equation are complex (i.e. D1, 2 = a ± ib), the general solution to the differential equation will be:

Calculation:

Given \(\frac{{{d^2}y}}{{d{x^2}}} + p\frac{{dy}}{{dx}} + qy = 0\) 

The given solution is:

\(y = {c_1}{e^{ - x}} + {c_2}{e^{ - 3x}}\)    ---(1)

The general solution of differential equation with unequal roots is given by:

\(y = {c_1}{e^{mx}} + {c_2}{e^{nx}}\)    ---(2)

Here m and n are the roots of ordinary differential equation

Comparing equation (1) and (2), we get:

m = -1 and n = -3

Sum of roots will be:

m + n = -p

-1 – 3 = -p

p = 4

And product of roots is mn = q, i.e.

(-1)(-3) = q