Solve the initial value problem:
25
𝑥
″
+
20
𝑥
′
+
229
𝑥
=
0
,
𝑥
(
0
)
=
2
,
𝑥
′
(
0
)
=
−
2.

Question

Solve the initial value problem:             
25
𝑥
″
+
20
𝑥
′
+
229
𝑥
=
0
,
𝑥
(
0
)
=
2
,
𝑥
′
(
0
)
=
−
2.

BlackTom AI · Accepted Answer

We start by identifying the differential equation and initial conditions given: 25 x'' + 20 x' + 229 x = 0, with x(0) = 2 and x'(0) = -2.

Step 1: Solve the characteristic equation. For a constant-coefficient ODE of the form 25 r^2 + 20 r + 229 = 0, compute the discriminant: Δ = 20^2 - 4·25·229 = 400 - 22900 = -22500, which is negative, indicating complex roots.

Step 2: Find the roots. The roots are r = [-20 ± sqrt(-22500)] / (2·25) = [-20 ± i·150] / 50 = -2/5 ± 3i.

Step 3: Write the general solution. With complex roots −2/5 ± 3i, the real-valued general solution is x(t) = e^{−(2/5)t} [A cos(3t) + B sin(3t)], where A and B are constants to be determined from initial conditions.

Step 4: Apply x(0) = 2. At t = 0, cos(0) = 1 and sin(0) = 0, so x(0) = e^{0} [A·1 + B·0] = A. Therefore A = 2.

Step 5: Compute x'(t) for the initial velocity. Differentiate x(t):
- x(t) = e^{−(2/5)t} [A cos(3t) + B sin(3t)].
- x'(t) = e^{−(2/5)t} [−(2/5)(A cos(3t) + B sin(3t)) + (−3A sin(3t) + 3B cos(3t))].
Evaluate at t = 0: cos(0) = 1, sin(0) = 0, so x'(0) = [−(2/5)(A) + 3B].

Step 6: Apply x'(0) = −2. With A = 2 from Step 4, we get −(2/5)·2 + 3B = −2, i.e. −4/5 + 3B = −2. Solve for B: 3B = −2 + 4/5 = −6/5, hence B = −2/5.

Step 7: Write the particular solution. Substituting A = 2 and B = −2/5 into the general form: x(t) = e^{−(2/5)t} [2 cos(3t) − (2/5) sin(3t)]. This matches the provided expression: x(t) = (2 cos 3t − (2/5) sin 3t) e^{−2t/5}.

Note: In the given data, the answer options field is empty, so there is no list of choices to evaluate. The derived result stands as the solution to the initial value problem based on standard methods (characteristic equation for linear constant-coefficient ODE, followed by applying initial conditions).

SMAT011 Weekly Quiz 9 |LA009

Solve the initial value problem: 25 𝑥 ″ + 20 𝑥 ′ + 229 𝑥 = 0 , 𝑥 ( 0 ) = 2 , 𝑥 ′ ( 0 ) = − 2.

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