Math Problem Statement

(D^2 + 16)^2 y = 0 with initial conditions y(0) = -3, y'(0) = -8, y''(0) = 32, y'''(0) = 256, and using t as the independent variable.

Solution

We are solving the given differential equation:

(D2+16)2y=0,(D^2 + 16)^2 y = 0, with initial conditions: y(0)=3,y(0)=8,y(0)=32,y(0)=256.y(0) = -3, \quad y'(0) = -8, \quad y''(0) = 32, \quad y'''(0) = 256. and using tt as the independent variable.

Step 1: Rewrite the operator form

Let D=ddtD = \frac{d}{dt}, so the equation becomes: (D2+16)2y=0.(D^2 + 16)^2 y = 0. Expanding, this represents a fourth-order differential equation.

Step 2: Solve the characteristic equation

The characteristic equation of D2+16D^2 + 16 is: r2+16=0    r=±4i.r^2 + 16 = 0 \implies r = \pm 4i. Since r=±4ir = \pm 4i are roots with multiplicity 2, the general solution to y(t)y(t) is: y(t)=(C1+C2t)cos(4t)+(C3+C4t)sin(4t),y(t) = (C_1 + C_2 t) \cos(4t) + (C_3 + C_4 t) \sin(4t), where C1,C2,C3,C4C_1, C_2, C_3, C_4 are constants determined by initial conditions.

Step 3: Compute derivatives of y(t)y(t)

  1. First derivative: y(t)=4(C1+C2t)sin(4t)+4(C3+C4t)cos(4t)+C2cos(4t)+C4sin(4t).y'(t) = -4(C_1 + C_2 t) \sin(4t) + 4(C_3 + C_4 t) \cos(4t) + C_2 \cos(4t) + C_4 \sin(4t).
  2. Second derivative: y(t)=16(C1+C2t)cos(4t)16(C3+C4t)sin(4t)8C2sin(4t)+8C4cos(4t).y''(t) = -16(C_1 + C_2 t) \cos(4t) - 16(C_3 + C_4 t) \sin(4t) - 8 C_2 \sin(4t) + 8 C_4 \cos(4t).
  3. Third derivative: y(t)=64(C1+C2t)sin(4t)64(C3+C4t)cos(4t)12C2cos(4t)12C4sin(4t).y'''(t) = 64(C_1 + C_2 t) \sin(4t) - 64(C_3 + C_4 t) \cos(4t) - 12C_2 \cos(4t) - 12C_4 \sin(4t).

Step 4: Apply initial conditions

  1. y(0)=3y(0) = -3: C1=3.C_1 = -3.
  2. y(0)=8y'(0) = -8: y(0)=4C1+4C3+C2=8    4(3)+4C3+C2=8    C2+4C3=20.y'(0) = -4C_1 + 4C_3 + C_2 = -8 \implies -4(-3) + 4C_3 + C_2 = -8 \implies C_2 + 4C_3 = -20.
  3. y(0)=32y''(0) = 32: y(0)=16C116C3+8C4=32    16(3)16C3+8C4=32    16C3+8C4=16.y''(0) = -16C_1 - 16C_3 + 8C_4 = 32 \implies -16(-3) - 16C_3 + 8C_4 = 32 \implies -16C_3 + 8C_4 = -16.
  4. y(0)=256y'''(0) = 256: y(0)=64C164C312C212C4=256    64(3)64C312C212C4=256    19264C312C212C4=256.y'''(0) = 64C_1 - 64C_3 - 12C_2 - 12C_4 = 256 \implies 64(-3) - 64C_3 - 12C_2 - 12C_4 = 256 \implies -192 - 64C_3 - 12C_2 - 12C_4 = 256.

Step 5: Solve for constants

From the equations:

  1. C1=3C_1 = -3,
  2. C2+4C3=20C_2 + 4C_3 = -20,
  3. 16C3+8C4=16    2C3+C4=2-16C_3 + 8C_4 = -16 \implies -2C_3 + C_4 = -2,
  4. 64C312C212C4=448.-64C_3 - 12C_2 - 12C_4 = 448.

Solve this system to find C2,C3,C4C_2, C_3, C_4.

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Math Problem Analysis

Mathematical Concepts

Differential Equations
Characteristic Equation
Initial Value Problem

Formulas

Characteristic equation r^2 + 16 = 0
General solution: y(t) = (C1 + C2 t) cos(4t) + (C3 + C4 t) sin(4t)

Theorems

Superposition Principle
Complex Root Behavior in Differential Equations

Suitable Grade Level

College Level (Calculus II or Differential Equations)