Math Problem Statement
solve
Solution
Let's solve the given problem step by step from the Laplace transform part onward.
Problem:
The differential equation is:
We already have the Laplace transform taken in the uploaded file:
Substituting the initial conditions and , the equation simplifies to:
Reorganizing:
Thus:
Step 1: Take the inverse Laplace transform term by term
First term:
This is the Laplace transform of :
Second term:
We need to use partial fraction decomposition to split this into simpler terms.
Let:
Multiply through by :
Expanding both terms:
Group terms by powers of :
Equating coefficients with :
- Coefficient of :
- Coefficient of :
- Coefficient of : (already known)
- Constant term:
Substitute into :
Thus, the partial fraction decomposition is:
Now, take the inverse Laplace transform of each term.
- is ,
- is .
Step 2: Combine the results
The solution is:
Simplify:
Final Answer:
Would you like me to explain any specific step in greater detail? Let me know!
Here are 5 related questions you might find helpful:
- What is the Laplace transform and how is it used to solve differential equations?
- How do we perform partial fraction decomposition for more complex fractions?
- How does the inverse Laplace transform work for trigonometric functions?
- Can we solve this differential equation using a different method, like undetermined coefficients?
- How does changing the initial conditions affect the solution?
Tip: Always double-check your partial fraction decomposition by substituting it back into the original equation to ensure correctness!
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Math Problem Analysis
Mathematical Concepts
Differential Equations
Laplace Transform
Inverse Laplace Transform
Partial Fraction Decomposition
Trigonometric Functions
Formulas
Laplace Transform: \(\mathcal{L}\{f(t)\} = F(s)\)
Inverse Laplace Transform: \(\mathcal{L}^{-1}\{F(s)\} = f(t)\)
Partial Fraction Decomposition
Theorems
Superposition Principle for Linear Systems
Laplace Transform Theorem for Derivatives
Suitable Grade Level
Undergraduate Mathematics or Advanced High School (Grade 12)
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