Math Problem Statement
y''+16y=0, y(0)=2, y'(0)=-2
Solution
To solve the differential equation
with the initial conditions and , we proceed as follows:
Step 1: Solve the Characteristic Equation
The given equation is a second-order linear homogeneous differential equation with constant coefficients. The characteristic equation is obtained by assuming a solution of the form , leading to:
Solving this, we get:
So, the general solution to the differential equation is:
Step 2: Apply Initial Conditions
-
Using :
Substitute :
So, .
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Using :
First, compute the derivative of :
Substitute :
So, .
Step 3: Write the Particular Solution
The particular solution that satisfies the given initial conditions is:
Final Answer:
Would you like further details or have any questions?
Related Questions:
- How do you solve second-order linear homogeneous differential equations with different characteristic roots?
- What is the role of initial conditions in determining a unique solution for a differential equation?
- How would the solution change if the initial conditions were different?
- What is the significance of complex roots in the characteristic equation?
- Can you explain the physical meaning of cosine and sine terms in solutions of differential equations?
Tip: When dealing with trigonometric solutions, use initial conditions to determine coefficients accurately, ensuring the solution fits the specific problem context.
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Math Problem Analysis
Mathematical Concepts
Differential Equations
Second-Order Linear Equations
Initial Value Problems
Formulas
Characteristic Equation for Second-Order Linear Equations
Theorems
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Suitable Grade Level
Advanced Undergraduate
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