Chapter 8: Problem 12
\(y^{\prime \prime \prime}+y^{\prime}=0, y(0)=y^{\prime}(0)=0, y^{\prime \prime}(0)=1\)
Short Answer
Expert verified
The solution is \( y(t) = -\text{cos}(t)\).
Step by step solution
01
Identify the Type of Differential Equation
This differential equation is a third-order linear homogeneous differential equation: \[y^{\frac{d^3}{dx^3}} + y^{\frac{d}{dx}} = 0\].
02
Find the Characteristic Equation
Rewrite the differential equation in terms of the characteristic equation, assuming a solution of the form \(y = e^{rt}\): \[r^3 + r = 0\].
03
Solve the Characteristic Equation
Solve the equation \(r^3 + r = 0\) by factoring: \[r(r^2 + 1) = 0\]. This gives the roots: \[r = 0, r = i, r = -i\].
04
Write the General Solution
Combine the solutions based on the roots found: \[ y(t) = C_1 + C_2 \text{cos}(t) + C_3 \text{sin}(t)\].
05
Apply Initial Conditions
Use the initial conditions to find the constants: Firstly, \(y(0) = 0\): \[0 = C_1 + C_2 \text{cos}(0) + C_3 \text{sin}(0)\]\[0 = C_1 + C_2\]. From the next initial condition, \(y^{\frac{d}{dt}}(0) = 0\): \[0 = 0 + (-C_2 \text{sin}(0) + C_3 \text{cos}(0))\]\[0 = C_3\]. With \(y''(0) = 1\): \[1 = 0 - C_2 \text{cos}(0)\]\[1 = -C_2\]\[C_2 = -1\].
06
Derive the Final Solution
With constants substituted in the general solution: \[ y(t) = 0 - \text{cos}(t) + 0 \text{sin}(t)\] Therefore, \[ y(t) = -\text{cos}(t)\].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Characteristic Equation
A characteristic equation is used to find solutions to linear differential equations with constant coefficients. In this problem, we start by rewriting the third-order differential equation in a simpler form. Assume the solution is of the form \(y = e^{rt}\), where \(r\) represents the roots of the characteristic equation. The given differential equation \(y^{\frac{d^3}{dx^3}} + y^{\frac{d}{dx}} = 0\) can be transformed into its characteristic form: \[r^3 + r = 0\]. This results from differentiating \(e^{rt}\). By solving this equation, you get the characteristic roots.
Initial Conditions
Initial conditions help determine the specific solution to a differential equation that fits particular criteria. For this problem, we are provided with: \(y(0)=0\), \(y^{\frac{d}{dt}}(0)=0\), and \(y''(0)=1\). Initial conditions allow you to find the constants in the general solution. For instance, \(y(0) = 0\) helps us find that: \[0 = C_1 + C_2 \text{cos}(0) + C_3 \text{sin}(0)\] This simplifies to \(C_1 + C_2 = 0\). By considering \(y^{\frac{d}{dt}}(0) = 0\), we get another equation: \[0 = -C_2 \text{sin}(0) + C_3 \text{cos}(0)\] This leads to \(C_3 = 0\). Finally, \(y^{\frac{d^2}{dt^2}}(0) = 1\) gives us: \[1 = -C_2 \text{cos}(0)\] Hence, \(C_2 = -1\). These conditions help build the unique solution to our differential equation.
General Solution
The general solution for a differential equation consists of a combination of solutions based on the roots of its characteristic equation. In our example, having roots \(r = 0\), \(r = i\), and \(r = -i\) leads to the general solution: \[ y(t) = C_1 + C_2 \text{cos}(t) + C_3 \text{sin}(t)\]. Using the initial conditions, we solved for the constants: \(C_1\), \(C_2\), and \(C_3\). Specifically, found that \(C_1 = 0\) (from \(C_1 + C_2 = 0\)), \(C_3 = 0\) (from \(C_3 = 0\)), and \(C_2 = -1\) (from \(C_2 = -1\)). After substituting these constants back into our general solution, we obtain the final solution: \(y(t) = -\text{cos}(t)\). This solution fulfills the original differential equation as well as the given initial conditions.