The general solution of the differential equation with the given initial conditions is x(t) = (1/2) * e^(-t) + (1/3) * e^(-3t).
Finding the general solution
Characteristic equation:
First, rewrite the differential equation in terms of the characteristic polynomial:
m^2 + 7m + 3 = 0
Factor the polynomial:
(m+1)(m+3) = 0
Therefore, the characteristic roots are m=-1, m=-3.
General solution:
The general solution for a homogeneous linear differential equation with constant coefficients is:
x(t) = c1 * e^(m1t) + c2 * e^(m2t)
where c1 and c2 are arbitrary constants and m1 and m2 are the characteristic roots.
Applying the characteristic roots:
Using the characteristic roots we found, the general solution becomes:
x(t) = c1 * e^(-t) + c2 * e^(-3t)
Finding the constants using initial conditions
Applying initial conditions:
We are given that x(2) = 1 and x'(2) = 2. Let's use these conditions to find the values of c1 and c2.
For x(2) = 1:
1 = c1 * e^(-2) + c2 * e^(-6)
For x'(2) = 2:
2 = -c1 * e^(-2) - 3c2 * e^(-6)
Solving the system of equations:
We can rewrite the system of equations as a matrix equation:
| e^(-2) e^(-6) | | c1 | | 1 |
| -e^(-2) -3e^(-6) | | c2 | | 2 |
Solving this system, we find:
c1 = 1/2 and c2 = 1/3
Final solution
Substituting the constants:
Substituting the values of c1 and c2 into the general solution, we get the final solution
x(t) = (1/2) * e^(-t) + (1/3) * e^(-3t)
Therefore, the general solution of the differential equation with the given initial conditions is:
x(t) = (1/2) * e^(-t) + (1/3) * e^(-3t)