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Magazine Chapter 17: Problem 21
Solve the initial-value problem. $$ y^{\prime \prime}-6 y^{\prime}+10 y=0, \quad y(0)=2, \quad y^{\prime}(0)=3 $$
Short Answer
Expert verified
The solution is \( y(t) = e^{3t}(2\cos(t) - 3\sin(t)) \).
Step by step solution
01
Identify and Classify
This is a second-order linear homogeneous differential equation with constant coefficients. The characteristic equation can be derived from the standard form: \( ar^2 + br + c = 0 \) where \( a = 1, b = -6, \) and \( c = 10 \). Form the characteristic equation: \( r^2 - 6r + 10 = 0 \).
02
Solve Characteristic Equation
Apply the quadratic formula to solve the characteristic equation, \( r^2 - 6r + 10 = 0 \): \[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{6 \pm \sqrt{36 - 40}}{2} = \frac{6 \pm \sqrt{-4}}{2} = \frac{6 \pm 2i}{2} = 3 \pm i \].The roots are complex: \( 3 \pm i \).
03
Form the General Solution
Since the roots are complex, the general solution of the differential equation is \( y(t) = e^{\alpha t}(C_1 \cos(\beta t) + C_2 \sin(\beta t)) \),where \( \alpha = 3 \) and \( \beta = 1 \). Hence, the general solution is \( y(t) = e^{3t}(C_1 \cos(t) + C_2 \sin(t)) \).
04
Apply Initial Conditions
Use the initial conditions \( y(0) = 2 \) and \( y'(0) = 3 \) to find \( C_1 \) and \( C_2 \). Start by computing the derivatives: \( y(0) = e^{0}(C_1 \cos(0) + C_2 \sin(0)) = C_1 = 2 \), so \( C_1 = 2 \).
05
Compute Derivative and Solve
Find the first derivative of \( y(t) \): \( y'(t) = e^{3t}(C_1 \cos(t) + C_2 \sin(t))' = e^{3t}[3(C_1 \cos(t) + C_2 \sin(t)) + C_1(-\sin(t)) + C_2\cos(t)] \)Substituting \( t = 0 \) gives: \( y'(0) = 3C_1 + C_2 = 3 \times 2 + C_2 = 3 \), solving this gives \( C_2 = -3 \).
06
Write the Particular Solution
Substitute \( C_1 = 2 \) and \( C_2 = -3 \) back into the general solution: \( y(t) = e^{3t}(2\cos(t) - 3\sin(t)) \).This is the solution satisfying the given initial conditions.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Initial-Value Problem
An initial-value problem in differential equations is a problem where you are asked to find a function that satisfies a differential equation and also meets specific conditions at the start, known as initial conditions. In our given problem, you have a second-order differential equation:
- The equation: \( y'' - 6y' + 10y = 0 \)
- Initial conditions given: \( y(0) = 2 \) and \( y'(0) = 3 \)
Homogeneous Equations
A homogeneous differential equation is one where every term depends on the function or its derivatives, and there are no standalone constant terms. For example, the equation \( y'' - 6y' + 10y = 0 \) is homogeneous. Some important points to note are:
- The term "homogeneous" signifies that if you multiply the function by any constant, the equation remains balanced.
- Solutions are typically formed by solving the characteristic equation derived from the differential equation.
Characteristic Equation
The characteristic equation is a key step in solving linear differential equations, particularly those with constant coefficients. For our differential equation, setting up the characteristic equation involves:
- Using the form \( ar^2 + br + c = 0 \) for a second-order equation, where \( a \), \( b \), and \( c \) are constants from the original differential equation.
- Resulting characteristic equation for our problem: \( r^2 - 6r + 10 = 0 \).
Complex Roots
The solutions to the characteristic equation can often be complex numbers. In our problem, the equation \( r^2 - 6r + 10 = 0 \) yielded roots \( 3 \pm i \). For complex roots \( \alpha \pm \beta i \), the solution to the differential equation takes a specific form:
- General solution: \( y(t) = e^{\alpha t}(C_1 \cos(\beta t) + C_2 \sin(\beta t)) \).
- Here, \( \alpha = 3 \) and \( \beta = 1 \) using the roots from our problem.
