Question

Consider the initial value problem: y′′+36y=e−ty′′+36y=e−t, y(0)=y0y(0)=y0, y′(0)=y′0y′(0)=y0′. Suppose we know that y(t)→0y(t)→0 as t→[infinity]t→[infinity]. Determine the solution and the initial conditions.

Answer 1

Answer:

This differential equation can be solved with the method of the undetermined coefficients

The general solution is yg = yh + yp

where yh is the solution to the homogeneous equation and yp is a particular solution

first of all, we calculate the homogeneous solution

y'' + 36y = 0

with the characteristic polynomial

$m^{2} - 36 = 0$

m = ± 6i

here we have imaginary solutions, hence:

yh = C1*cos(6t) + C2*sin(6t)

With C1 and C2 constants that we have to compute with the initial conditions. For the calculation of yp we assume that one solution is yp = C3*e^(-t)

by calculating the derivatives and replacing:

yp' = -C3*e^(-t)

yp'' = C3*e^(-t)

yp'' + 36yp = e^(-t)

C3*e^(-t) + 36*C3*e^(-t) = e^(-t)

37C3 = 1   -->   C3 = 1/37

Hence, the solution is

y = yg = yh + yp = C1*cos(6t) + C2*sin(6t) + (1/37)e^(-t)

By using the initial condition we have

y(0) = C1*cos(0) + C2*sin(0) + (1/37)e^(0) = C1 + 1/37 = 0  -->  C1 = -1/37

y'(0) = -C1*sin(0) + C2*cos(0) - (1/37)e^(0) = C2 - 1/37 = 0  -->  C2= 1/37

Thus, finally

y = (1/37)( cos(6t) + sin(6t) + e^(-t) )