Math Problem Statement
𝑦‴+𝑦″−16𝑦′+20𝑦=0;𝑦(0)=0,𝑦′(0)=1,𝑦″(0)=3.
Solution
We are given the third-order linear differential equation:
with the initial conditions:
Step 1: Solve the characteristic equation
The first step in solving a linear differential equation with constant coefficients is to write down the characteristic equation. For this equation, the characteristic equation is obtained by replacing with , with , with , and with :
Step 2: Find the roots of the characteristic equation
We will now solve the cubic equation:
Let's use the Rational Root Theorem to check for possible rational roots. The possible rational roots are factors of the constant term (20) divided by factors of the leading coefficient (1), so the possible rational roots are .
Let's try some values. Testing :
Thus, is a root of the characteristic equation.
Step 3: Factor the cubic polynomial
Since is a root, we can factor as:
Now, solve the quadratic factor using the quadratic formula:
where , , and :
So the roots are:
Step 4: General solution
The general solution to the differential equation is:
Simplifying:
Step 5: Apply initial conditions
Now we will use the initial conditions to solve for the constants , , and .
- Condition :
- Condition :
First, find :
- Condition :
Next, find :
Step 6: Solve the system of equations
We now solve the system of three equations:
Solving this system:
- From the first equation, .
- Substitute into the other two equations:
Now substitute into :
Substitute and into the third equation:
[ -\frac{21}{7} = 3 \implies c_3 . \
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Math Problem Analysis
Mathematical Concepts
Differential Equations
Characteristic Equation
Cubic Equations
Initial Conditions
Linear Differential Equations with Constant Coefficients
Formulas
Characteristic Equation: r^3 + r^2 - 16r + 20 = 0
Quadratic Formula: r = (-b ± √(b^2 - 4ac)) / 2a
Theorems
Rational Root Theorem
Quadratic Formula
Suitable Grade Level
Grades 11-12
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