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Magazine Chapter 4: Problem 66
Solve the given differential equation subject to the indicated initial conditions. $$ y^{\prime \prime}+y^{\prime}=x, \quad y(0)=1, y^{\prime}(0)=0 $$
Short Answer
Expert verified
The solution is \(y = e^{-x} + x\).
Step by step solution
01
Identify the Type of Equation
The given differential equation is \(y'' + y' = x\). This is a non-homogeneous linear second-order differential equation with constant coefficients.
02
Solve the Homogeneous Equation
Consider the associated homogeneous equation \(y'' + y' = 0\). The characteristic equation is \(r^2 + r = 0\). Factoring gives \(r(r+1) = 0\), which has roots \(r = 0\) and \(r = -1\). Therefore, the general solution for the homogeneous equation is \(y_h = C_1 + C_2e^{-x}\).
03
Find Particular Solution
To find a particular solution, \(y_p\), to \(y'' + y' = x\), assume a form that mimics the non-homogeneous part. A suitable choice is \(y_p = Ax + B\). Find its derivative: \(y_p' = A\) and \(y_p'' = 0\). Substituting into the original equation gives \(0 + A = x\). Thus, \(A = 1\) and we solve for \(B\) later if needed. Thus, \(y_p = x + B\).
04
Combine Solutions
The general solution to the differential equation is the sum of the homogeneous and particular solutions: \(y = y_h + y_p = C_1 + C_2e^{-x} + x + B\). Set \(B = 0\) simply as it is unnecessary to find it if initial conditions will help determine the remaining constants.
05
Apply Initial Conditions
Using \(y(0) = 1\), substitute into \(y = C_1 + C_2e^0 + 0 + 0\), so \(y(0) = C_1 + C_2 = 1\). For \(y'(0) = 0\), calculate \(y' = C_2(-e^{-x}) + 1\) and substitute \(y'(0) = -C_2 + 1 = 0\), thus \(C_2 = 1\). Substituting \(C_2 = 1\) into \(y(0) = C_1 + C_2 = 1\) gives \(C_1 = 0\).
06
Write the Final Solution
The final solution after applying the initial conditions is \(y = e^{-x} + x\). This solution satisfies both the differential equation and the initial conditions given in the problem.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Initial Conditions
Initial conditions are crucial in finding a unique solution to a differential equation. They specify the value of the solution and its derivatives at a certain point, usually where the problem starts. In our exercise, the initial conditions given are:
- \( y(0) = 1 \), which tells us the value of the function \( y \) at \( x = 0 \).
- \( y'(0) = 0 \), which tells us the value of the first derivative \( y' \) at \( x = 0 \).
Second-Order Linear Differential Equation
A second-order linear differential equation involves derivatives up to the second order. It typically has the form:\[ a(x)y'' + b(x)y' + c(x)y = g(x) \]In this exercise, the equation is:\[ y'' + y' = x \]This particular equation is linear and has constant coefficients, meaning the coefficients of \( y'' \) and \( y' \) do not depend on \( x \). The equation is called non-homogeneous because of the presence of the term \( x \) on the right side, which doesn't contain the function \( y \) or its derivatives.
Characteristic Equation
The characteristic equation is a key tool when solving a homogeneous linear differential equation with constant coefficients. For a homogeneous equation such as:\[ y'' + y' = 0 \]the characteristic equation is derived by replacing \( y'' \) with \( r^2 \) and \( y' \) with \( r \), resulting in:\[ r^2 + r = 0 \]This equation can be factored as:\[ r(r+1) = 0 \]The roots of this equation, \( r = 0 \) and \( r = -1 \), identify the form of the solution to the homogeneous part. Each root provides a component of the solution. Therefore, the general solution for the homogeneous differential equation is:\[ y_h = C_1 + C_2e^{-x} \]
Homogeneous Solution
A homogeneous solution \( y_h \) addresses the part of the differential equation without any non-zero terms or inhomogeneities - essentially solving \( y'' + y' = 0 \). This equation neglects any forcing terms (i.e., \( x \)).The general solution is formed by using the roots from the characteristic equation:\[ y_h = C_1 + C_2e^{-x} \]These terms account for the natural response of the system described by the differential equation. Each part of the homogeneous solution can be adjusted using constants \( C_1 \) and \( C_2 \), which we find using the initial conditions. The homogeneous solution is always part of the complete solution to the differential equation.
