SHAPE: y "+ a1y '+ a0y = 0 (a0, a1 constants)
Ex: y =
So y '=
and y "=
Substituting in the given equation:
() = 0
0 for every x, so we should have = 0, which is an equation of the second degree in the variable , call characteristic equation.
The solution of the linear differential equation will depend on the roots 1 and 2.
- 1, 2 real and distinct numbers C1 and C2 are particular EDO solutions and the general solution is y = C1 + C2
- 1 = 2 = (real and equal numbers) EDO's overall solution is y = C1 + C2x
- 1 = a + bi, 2 = a - bi (conjugated complexes: a, b real) the general solution is y = C1 + C2
Ex: y "- 2y '- 15y = 0
Characteristic equation: - 2 - 15 = 0 whose roots are: 1 = 5, 2= -3
General solution: y =Next: Linear Differential Equations of Order N