Chapter 6: Problem 41
Solve each differential equation and initial condition and verify that your answer satisfies both the differential equation and the initial condition. $$ \left\\{\begin{array}{l} y^{\prime}=2 x y^{2} \\ y(0)=1 \end{array}\right. $$
Short Answer
Expert verified
The solution is \( y = -\frac{1}{x^2 - 1} \). It satisfies the differential equation and initial condition.
Step by step solution
01
Identify the Differential Equation Type
The given differential equation is \( y' = 2xy^2 \). This is a first-order ordinary differential equation, and it can be identified as a separable differential equation because it can be written as a product of functions, each depending only on one variable.
02
Separate Variables
We can separate the variables by rewriting the equation as \( \frac{dy}{y^2} = 2x \, dx \). This allows us to integrate both sides with respect to their respective variables.
03
Integrate Both Sides
Integrate the left side with respect to \( y \) and the right side with respect to \( x \). The integration gives:\[ \int \frac{dy}{y^2} = \int 2x \, dx \]\[-\frac{1}{y} = x^2 + C \] (where \( C \) is the constant of integration).
04
Solve for y
Solve the equation \(-\frac{1}{y} = x^2 + C \) for \( y \):\[ y = -\frac{1}{x^2 + C} \]
05
Apply Initial Condition
Use the initial condition \( y(0) = 1 \) to find the constant \( C \).Substitute \( x = 0 \) and \( y = 1 \) into \( y = -\frac{1}{x^2 + C} \):\[ 1 = -\frac{1}{0 + C} \]So, \( C = -1 \).
06
Write the Particular Solution
With the constant \( C = -1 \), the particular solution of the differential equation is:\[ y = -\frac{1}{x^2 - 1} \]
07
Verify the Solution
Verify that the solution satisfies both the differential equation and the initial condition. Differentiate the solution with respect to \( x \):\( y = -\frac{1}{x^2 - 1} \) implies \( y' = \frac{ 2x }{(x^2 - 1)^2} \).Plug \( y = -\frac{1}{x^2 - 1} \) into the differential equation \( y' = 2xy^2 \):\[ \frac{2x}{(x^2 - 1)^2} = 2x \left(-\frac{1}{x^2 - 1}\right)^2 \]Both sides are equal, confirming the solution is correct.Also, \( y(0) = -\frac{1}{-1} = 1 \), which satisfies the initial condition.
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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 ordinary differential equations that can be split into two separate integrals. They are relatively easier to solve because of their structure. To identify a separable differential equation, you should be able to write it in the form:
- \( \frac{dy}{dx} = g(x)h(y) \)
- This can be rewritten as \( \frac{1}{h(y)} \, dy = g(x) \, dx \)
Initial Conditions
Initial conditions are constraints that allow us to find a particular solution out of the infinite possible solutions of a differential equation. They provide specific values at a given point, ensuring the solution meets specific criteria. For example, with the initial condition:
- If given \( y(0) = 1 \), you substitute \( x = 0 \) into your general solution to find the unknown constant, C.
- This process picks out a unique solution from the general solution set that satisfies this condition.
Ordinary Differential Equations
Ordinary differential equations (ODEs) are equations involving derivatives of a function with respect to one variable. They are called 'ordinary' because they involve derivatives with respect to a single variable, as opposed to partial differential equations which involve multiple variables. A first-order ODE contains functions and their first derivatives, marked often as \( y' \). In our earlier example, we worked with the first-order ODE \( y' = 2x y^2 \). Key properties of ODEs include:
- They model constant or predictable rates of change in dynamic systems.
- They are essential tools in mathematics and its applications, including physics, economics, and biology.