Math Problem Statement

Find the general solution of the differential equation y'' - 2y' - 8y = -16t + 16t^2.

Solution

To solve the differential equation

y2y8y=16t+16t2,y'' - 2y' - 8y = -16t + 16t^2,

we'll approach it by finding the general solution, which consists of two parts: the complementary solution ycy_c and the particular solution ypy_p.

Step 1: Find the Complementary Solution ycy_c

The complementary solution solves the homogeneous equation: y2y8y=0.y'' - 2y' - 8y = 0.

  1. Start by finding the characteristic equation: r22r8=0.r^2 - 2r - 8 = 0.
  2. Solve this quadratic equation using the quadratic formula: r=2±(2)2+482=2±62.r = \frac{2 \pm \sqrt{(-2)^2 + 4 \cdot 8}}{2} = \frac{2 \pm 6}{2}. This gives roots r=4r = 4 and r=2r = -2.

Since both roots are real and distinct, the complementary solution is: yc=c1e4t+c2e2t.y_c = c_1 e^{4t} + c_2 e^{-2t}.

Step 2: Find the Particular Solution ypy_p

Since the non-homogeneous part is 16t+16t2-16t + 16t^2, a polynomial of degree 2, we assume a particular solution of the form: yp=At2+Bt+C.y_p = At^2 + Bt + C.

  1. Find ypy_p' and ypy_p'':

    • yp=2At+By_p' = 2At + B,
    • yp=2Ay_p'' = 2A.
  2. Substitute ypy_p, ypy_p', and ypy_p'' into the differential equation: 2A2(2At+B)8(At2+Bt+C)=16t+16t2.2A - 2(2At + B) - 8(At^2 + Bt + C) = -16t + 16t^2.

  3. Collect terms and equate coefficients to solve for AA, BB, and CC.

Step

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Math Problem Analysis

Mathematical Concepts

Second-order linear differential equations
Homogeneous and non-homogeneous equations
Characteristic equation
Particular solution using polynomial guess

Formulas

y'' - 2y' - 8y = 0 (Homogeneous equation)
Characteristic equation: r^2 - 2r - 8 = 0
Quadratic equation formula: r = (-b ± √(b^2 - 4ac)) / 2a
Particular solution form: At^2 + Bt + C

Theorems

Superposition principle
Method of undetermined coefficients

Suitable Grade Level

Grades 11-12 / College Level