Math Problem Statement
Solve the given initial value problem.
y double prime plus 5 y primey′′+5y′equals=0;
y(0)equals=negative 4−4,
y prime left parenthesis 0 right parenthesisy′(0)equals=negative 5−5
Question content area bottom
Part 1
The solution is
y(t)equals=enter your response here.
Solution
We are given the second-order linear differential equation:
with initial conditions:
Step 1: Solve the homogeneous equation
The given differential equation is:
This is a linear homogeneous differential equation with constant coefficients. The characteristic equation associated with this differential equation is:
We can factor this equation:
So, the roots are:
Thus, the general solution to the differential equation is:
Since , we simplify this to:
Step 2: Use the initial conditions to determine and
- Apply the first initial condition :
Substitute into the solution:
Thus, we get the equation:
C_1 + C_2 = -4. \tag{1}
- Apply the second initial condition :
First, find :
Substitute :
So, we get:
C_2 = 1. \tag{2}
Step 3: Solve for
Substitute into equation (1):
which gives:
Step 4: Write the final solution
Now that we have and , the solution to the differential equation is:
Final Answer:
Would you like further explanation on any part of the solution? Or do you have any other questions?
Related Questions:
- What is the behavior of the solution as ?
- How would the solution change if the equation had a different constant in front of ?
- Can this method be applied to higher-order linear differential equations?
- How do you solve non-homogeneous differential equations with similar methods?
- What is the significance of the initial conditions in determining the constants and ?
Tip: Always check if the roots of the characteristic equation are real or complex, as this determines the form of the solution (exponential vs. sinusoidal).
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Math Problem Analysis
Mathematical Concepts
Differential Equations
Initial Value Problems
Homogeneous Linear Differential Equations
Characteristic Equations
Formulas
Characteristic Equation: r^2 + 5r = 0
General Solution: y(t) = C1 e^(r1 t) + C2 e^(r2 t)
Derivative Rule: d/dt (C e^(at)) = a C e^(at)
Theorems
Theory of Linear Differential Equations with Constant Coefficients
Method of Solving Homogeneous Differential Equations
Suitable Grade Level
Undergraduate Calculus (Differential Equations)
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