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Magazine Chapter 2: Problem 10
In Problems 1-40 find the general solution of the given differential equation. State an interval on which the general solution is defined. $$ y^{\prime}=2 y+x^{2}+5 $$
Short Answer
Expert verified
The general solution is \(y(x) = C_2 e^{2x} - \frac{1}{2}x^2 - x - 3\) on \((-\infty, \infty)\).
Step by step solution
01
Identify the Type of Differential Equation
The given equation is \( y' = 2y + x^2 + 5 \). This is a first-order, linear non-homogeneous differential equation.
02
Solve the Homogeneous Equation
First, we solve the homogeneous part: \( y' = 2y \). This can be rewritten as \( \frac{dy}{dx} = 2y \). Separating variables gives \( \frac{dy}{y} = 2 \, dx \). Integrating both sides, we have \( \ln |y| = 2x + C_1 \). Exponentiating both sides gives \( y = C_2 e^{2x} \), where \( C_2 = e^{C_1} \).
03
Find a Particular Solution
For the non-homogeneous part \( y' = 2y + x^2 + 5 \), we use the method of undetermined coefficients. Assume a particular solution of the form \( y_p = Ax^2 + Bx + C \). Differentiating gives \( y_p' = 2Ax + B \). Substitute \( y_p \) and \( y_p' \) into the differential equation: \( 2Ax + B = 2(Ax^2 + Bx + C) + x^2 + 5 \). Equate coefficients to solve for \( A, B, \) and \( C \).
04
Solve for Coefficients
Comparing coefficients from the previous step gives:- Coefficient of \( x^2 \): \( 0 = 2A + 1 \) implies \( A = -\frac{1}{2} \).- Coefficient of \( x \): \( 2A = 2B \) implies \( B = -1 \).- Constant term: \( B = 2C + 5 \) implies \( C = -3 \).Thus, the particular solution is \( y_p = -\frac{1}{2}x^2 - x - 3 \).
05
Combine Homogeneous and Particular Solutions
The general solution is the sum of the homogeneous solution and the particular solution: \[ y(x) = C_2 e^{2x} - \frac{1}{2}x^2 - x - 3 \].
06
Determine the Interval for the Solution
The function \( y(x) = C_2 e^{2x} - \frac{1}{2}x^2 - x - 3 \) is composed of polynomial and exponential functions, which are defined for all real \( x \). Therefore, the general solution is valid on the interval \((-\infty, \infty)\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
First-Order Differential Equation
A first-order differential equation involves a function and its first derivative. In this context, "first-order" refers to the highest derivative, which is the first derivative, often denoted as \( y' \) or \( \frac{dy}{dx} \). These types of equations describe a wide range of phenomena and are fundamental in understanding changes across fields like physics, biology, and economics.
First-order differential equations can be categorized into various types, with linear and non-linear being the most common. The equation given in the exercise, \( y' = 2y + x^2 + 5 \), is a linear differential equation because it can be expressed in the standard linear form \( y' + P(x)y = g(x) \) where both \( P(x) \) and \( g(x) \) are functions of \( x \). Understanding the nature of the equation is the first step towards finding its solution.
First-order differential equations can be categorized into various types, with linear and non-linear being the most common. The equation given in the exercise, \( y' = 2y + x^2 + 5 \), is a linear differential equation because it can be expressed in the standard linear form \( y' + P(x)y = g(x) \) where both \( P(x) \) and \( g(x) \) are functions of \( x \). Understanding the nature of the equation is the first step towards finding its solution.
Linear Non-Homogeneous Equation
In many cases, differential equations are categorized as either homogeneous or non-homogeneous. A homogeneous linear differential equation consists only of terms involving the function and its derivatives, whereas a non-homogeneous equation has additional terms independent of the function.
The given equation, \( y' = 2y + x^2 + 5 \), is non-homogeneous due to the presence of \( x^2 + 5 \). This means that besides finding solutions that satisfy the derivative and multiplicative part involving \( y \), you need to account for this extra "driving force."
In practice, solving a non-homogeneous equation often involves finding both the homogeneous solution (solving \( y' = 2y \)) and a particular solution. These are then combined to form the general solution.
The given equation, \( y' = 2y + x^2 + 5 \), is non-homogeneous due to the presence of \( x^2 + 5 \). This means that besides finding solutions that satisfy the derivative and multiplicative part involving \( y \), you need to account for this extra "driving force."
In practice, solving a non-homogeneous equation often involves finding both the homogeneous solution (solving \( y' = 2y \)) and a particular solution. These are then combined to form the general solution.
Method of Undetermined Coefficients
The method of undetermined coefficients is a common strategy for finding particular solutions of non-homogeneous linear differential equations. This involves assuming a form for the particular solution and then determining the coefficients by substituting back into the original equation.
For the equation \( y' = 2y + x^2 + 5 \), assume a particular solution of the form \( y_p = Ax^2 + Bx + C \). Differentiate this assumed form to get \( y'_p = 2Ax + B \). Plugging these into the differential equation allows comparison of coefficients from both sides, providing equations to solve for \( A, B, \) and \( C \).
This method works well for differential equations where the non-homogeneous part is a polynomial, exponential, or trigonometric function. Simplifying the process of solving, it turns complex equations into straightforward algebraic challenges.
For the equation \( y' = 2y + x^2 + 5 \), assume a particular solution of the form \( y_p = Ax^2 + Bx + C \). Differentiate this assumed form to get \( y'_p = 2Ax + B \). Plugging these into the differential equation allows comparison of coefficients from both sides, providing equations to solve for \( A, B, \) and \( C \).
This method works well for differential equations where the non-homogeneous part is a polynomial, exponential, or trigonometric function. Simplifying the process of solving, it turns complex equations into straightforward algebraic challenges.
General Solution
The general solution of a differential equation is a formula that incorporates both the homogeneous solution and a particular solution, representing all possible solutions. It forms a complete set of solutions for the equation.
For example, after finding a homogeneous solution \( y_h = C_2 e^{2x} \) and a particular solution \( y_p = -\frac{1}{2}x^2 - x - 3 \), the general solution is their sum: \[ y(x) = C_2 e^{2x} - \frac{1}{2}x^2 - x - 3 \]
The validity of the general solution is established for all real numbers, making it a universal solution on the interval \( (-\infty, \infty) \). Such solutions are crucial for understanding dynamic systems across many real-world applications.
For example, after finding a homogeneous solution \( y_h = C_2 e^{2x} \) and a particular solution \( y_p = -\frac{1}{2}x^2 - x - 3 \), the general solution is their sum: \[ y(x) = C_2 e^{2x} - \frac{1}{2}x^2 - x - 3 \]
- The term \( C_2 e^{2x} \) represents all solutions to the homogeneous part, with \( C_2 \) being an arbitrary constant.
- The particular solution addresses the specific non-homogeneous part, setting the shape required to match the given equation.
The validity of the general solution is established for all real numbers, making it a universal solution on the interval \( (-\infty, \infty) \). Such solutions are crucial for understanding dynamic systems across many real-world applications.
