consider the differential equation
d^3x/dt^3 - 9(d^2x/dt^2)+ 27(dx/dt) -27x = c0s t +sin t + te^(3t)
a) show that characteristic equation of the differential equation is
(m-3)^3 =0
(b) Hence, find the general solution of the equation.
well, it's clear that
(m-3)^3 = m^3 - 9m^2 + 27m - 27
now just proceed as usual to solve this type of equation.
To solve the given differential equation, we will first find the characteristic equation associated with it, and then solve it to find the general solution.
(a) To find the characteristic equation, we substitute x = e^(mt) into the given differential equation. The derivatives of x with respect to t can be expressed in terms of m.
Substituting x = e^(mt) into the differential equation:
d^3x/dt^3 - 9(d^2x/dt^2) + 27(dx/dt) - 27x = c0s(t) + s1n(t) + te^(3t)
Taking the derivatives:
m^3e^(mt) - 9m^2e^(mt) + 27me^(mt) - 27e^(mt) = c0s(t) + s1n(t) + te^(3t)
Factoring out e^(mt):
e^(mt)(m^3 - 9m^2 + 27m - 27) = c0s(t) + s1n(t) + te^(3t)
Since e^(mt) is never zero, we can divide both sides of the equation by e^(mt):
m^3 - 9m^2 + 27m - 27 = c0s(t) + s1n(t) + te^(3t)
Comparing the resulting equation with (m-3)^3 = 0, we can see that they are identical. Therefore, the characteristic equation of the given differential equation is (m-3)^3 = 0.
(b) The characteristic equation (m-3)^3 = 0 has a triple root of m = 3. This means the general solution of the differential equation will have the form:
x(t) = (A + Bt + Ct^2)e^(3t)
where A, B, and C are arbitrary constants to be determined by any initial or boundary conditions given in the problem.
Thus, the general solution of the given differential equation is x(t) = (A + Bt + Ct^2)e^(3t).