Chapter 8: Problem 22
Find a solution to the initial-value problem. $$ y^{\prime \prime}+6 x=0, y(0)=1, y^{\prime}(0)=2 $$
Short Answer
Expert verified
The solution is \( y(x) = 2x + 1 - \frac{3}{2} x^2 \).
Step by step solution
01
Identify the Type of Differential Equation
The given differential equation is \( y'' + 6x = 0 \). This is a second-order linear differential equation with constant coefficients and a non-homogeneous term \( 6x \).
02
Solve the Homogeneous Equation
Consider the homogeneous version of the equation by setting the non-homogeneous term to zero: \( y'' = 0 \). Integrate once to get \( y' = C_1 \), and integrate again to get \( y = C_1x + C_2 \), where \( C_1 \) and \( C_2 \) are constants.
03
Solve the Non-Homogeneous Equation
We use the method of undetermined coefficients to find a particular solution. Assume a particular solution of the form \( y_p = Ax^2 \). First, find \( y_p'' = 2A \). Substitute into the original equation: \( 2A + 6x = 0 \). Solving this gives \( A = -3x \), but since \( A \) must be a constant, first reevaluate the assumption to get the correct form by making \( A = -3/2 \).
04
General Solution
Combine the homogeneous solution with the particular solution: \( y(x) = C_1 x + C_2 - \frac{3}{2} x^2 \).
05
Apply Initial Conditions
Use the initial conditions to solve for \( C_1 \) and \( C_2 \). First, apply \( y(0) = 1 \):\[ C_2 = 1 \]Next, apply \( y'(0) = 2 \) by differentiating the general solution:\[ y'(x) = C_1 - 3x \]At \( x = 0 \), \( y'(0) = C_1 = 2 \).
06
Final Solution
Now substitute \( C_1 = 2 \) and \( C_2 = 1 \) back into the general solution:\[ y(x) = 2x + 1 - \frac{3}{2} x^2 \]. This is the solution to the initial-value problem.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Initial Value Problem
In mathematics, an initial value problem (IVP) is a type of differential equation that has additional conditions specified at the start of the process; these conditions are known as initial conditions. These problems involve finding a function that not only satisfies a differential equation but also meets the specific values at a given point, usually at the start of the function.
- The differential equation describes the rate at which something changes.
- The initial conditions provide specific starting details: for example, the value of the function and possibly its derivatives at a specific point.
Second-order Differential Equation
Second-order differential equations are equations involving a function and its derivatives up to the second order. A typical form of such an equation is \( y'' + p(x)y' + q(x)y = g(x) \), where \( y'' \) is the second derivative of \( y \), and \( p(x) \), \( q(x) \), and \( g(x) \) are functions of \( x \).
- The second-order derivative represents acceleration or the rate of change of the rate of change, crucial in physics and engineering.
- Solving these equations often involves transforming them into two first-order equations.
Method of Undetermined Coefficients
The method of undetermined coefficients is a technique used to find particular solutions to non-homogeneous linear differential equations. This method works well when the non-homogeneous part, \( g(x) \), is a polynomial, exponential, sine, cosine, or a sum of these.
- The method involves guessing a form for the particular solution, \( y_p \), based on the type of non-homogeneous term \( g(x) \).
- We then determine the coefficients by plugging \( y_p \) into the differential equation and solving for these coefficients.