Math Problem Statement
Solution
To solve the differential equation
we'll approach it by finding the general solution, which consists of two parts: the complementary solution and the particular solution .
Step 1: Find the Complementary Solution
The complementary solution solves the homogeneous equation:
- Start by finding the characteristic equation:
- Solve this quadratic equation using the quadratic formula: This gives roots and .
Since both roots are real and distinct, the complementary solution is:
Step 2: Find the Particular Solution
Since the non-homogeneous part is , a polynomial of degree 2, we assume a particular solution of the form:
-
Find and :
- ,
- .
-
Substitute , , and into the differential equation:
-
Collect terms and equate coefficients to solve for , , and .
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
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