Chapter 16: Problem 3
Solve the given differential equation. $$y^{\prime}-x y^{2}=0, y(1)=2$$
Short Answer
Expert verified
The solution is \( y = \frac{2}{2 - x^2} \).
Step by step solution
01
Recognize the Type of Differential Equation
The given differential equation is \( y^{\prime} - xy^2 = 0 \). This is a first-order differential equation.
02
Rearrange the Equation into a Separable Form
To make the equation separable, rearrange the terms: \[y^{\prime} = x y^2\] This can be rewritten as:\[\frac{dy}{dx} = x y^2\] The goal is to separate variables \( y \) and \( x \) on opposite sides.
03
Separate Variables
Separate the equation into two integrable parts:\[\frac{1}{y^2} \, dy = x \, dx\] This achieves the separation of variables required to integrate both sides.
04
Integrate Both Sides
Integrate both sides:\[\int \frac{1}{y^2} \, dy = \int x \, dx\] The integrals are:\[-\frac{1}{y} = \frac{x^2}{2} + C\] where \( C \) is the constant of integration.
05
Solve for the Constant Using Initial Condition
Use the initial condition \( y(1) = 2 \) to find \( C \). Substitute \( x = 1 \) and \( y = 2 \):\[-\frac{1}{2} = \frac{1^2}{2} + C\]This simplifies to:\[-\frac{1}{2} = \frac{1}{2} + C\]Solve for \( C \):\[C = -1\]
06
Write the General Solution with the Specific Constant
Substitute \( C = -1 \) back into the integral solution:\[-\frac{1}{y} = \frac{x^2}{2} - 1\]Simplify to find \( y \):\[\frac{1}{y} = 1 - \frac{x^2}{2}\]\[y = \frac{1}{1 - \frac{x^2}{2}} = \frac{2}{2 - x^2}\]
07
State the Final Solution
The solution to the differential equation \( y^{\prime} - xy^2 = 0 \) given \( y(1)=2 \) is:\[y = \frac{2}{2 - x^2}\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Separable Differential Equations
Separable differential equations are a specific type of first-order differential equation. They allow us to separate the variables into two distinct groups, each with its integrable expression. This separation simplifies the process of finding a general solution to the differential equation.
In general, a differential equation is separable if it can be rearranged into the form:
By integrating each side with respect to its respective variable, we effectively isolate \( y \) in terms of \( x \) or vice versa, leading us closer to the solution.
In general, a differential equation is separable if it can be rearranged into the form:
- \( N(y)\, dy = M(x)\, dx \)
By integrating each side with respect to its respective variable, we effectively isolate \( y \) in terms of \( x \) or vice versa, leading us closer to the solution.
First-Order Differential Equations
A first-order differential equation involves the first derivative of a function, but no higher-order derivatives. The general form is \( y' = f(x, y) \), indicating how the derivative \( y' \) (or \( \frac{dy}{dx} \)) relates to \( x \) and \( y \). Knowing that a differential equation is of first order is important as it often dictates the methods used to solve it.
In the exercise \( y' - xy^2 = 0 \), we identify it as a first-order equation because it only contains the first derivative \( y' \). The absence of higher derivatives makes it a good candidate for specific methods like separation of variables. This understanding allows us to apply appropriate techniques, ensuring we solve the equation efficiently and accurately.
In the exercise \( y' - xy^2 = 0 \), we identify it as a first-order equation because it only contains the first derivative \( y' \). The absence of higher derivatives makes it a good candidate for specific methods like separation of variables. This understanding allows us to apply appropriate techniques, ensuring we solve the equation efficiently and accurately.
Integration Techniques
Integration is a fundamental tool required to solve separable and first-order differential equations. When you separate variables within a differential equation, you are often left with two sides that each need to be integrated. Two standard techniques of integration applied here are:
- Basic Antiderivatives: Recognizing the integral forms, such as \( \int x^n dx = \frac{x^{n+1}}{n+1} + C \), necessary for the integration of power functions.
- Initial Conditions: Utilization of specific conditions given in a problem (such as \( y(1) = 2 \)) to find the constant of integration, ensuring a unique solution rather than a general one.