A differential equation is given as x(n+2) + 3x(n+1) + 2x(n) = 0; x(0) = 0, x(1) = 1. The solution of this equation will be:
10) A differential equation is given as x(n+2) + 3x(n+1) + 2x(n) = 0; x(0) = 0, x(1) = 1. The solution of this equation will be:
Correct Answer: x(n) = (-1)n - (-2)n [Option ID = 2872]
Solution
Note: This is a Linear Difference Equation (Recurrence Relation). It is solved similarly to a second-order linear differential equation by finding characteristic roots.
1. Identify the Equation & Conditions
x(n+2) + 3x(n+1) + 2x(n) = 0
Initial Conditions: x(0) = 0, x(1) = 1
Initial Conditions: x(0) = 0, x(1) = 1
2. Form the Characteristic Equation
Assume a solution of the form x(n) = rn. Substituting this into the equation gives:
r2 + 3r + 2 = 0
3. Find the Roots
Factoring the quadratic equation:
(r + 1)(r + 2) = 0
r1 = -1, r2 = -2
r1 = -1, r2 = -2
4. Write the General Solution
For two distinct real roots, the general form is:
x(n) = A(r1)n + B(r2)n
x(n) = A(-1)n + B(-2)n
x(n) = A(-1)n + B(-2)n
5. Solve for Constants (A and B)
Using x(0) = 0:
0 = A(-1)0 + B(-2)0
0 = A + B ⇒ A = -B
0 = A + B ⇒ A = -B
Using x(1) = 1:
1 = A(-1)1 + B(-2)1
1 = -A - 2B
1 = -A - 2B
Substitute A = -B into the second equation:
1 = -(-B) - 2B
1 = B - 2B
1 = -B ⇒ B = -1
Since A = -B, A = 1
1 = B - 2B
1 = -B ⇒ B = -1
Since A = -B, A = 1
6. Final Solution
Substitute A and B back into the general equation:
x(n) = (1)(-1)n + (-1)(-2)n
x(n) = (-1)n - (-2)n
x(n) = (-1)n - (-2)n
This matches Option 4.