Question

Find the solution to the initial-value problem.\(y^{\prime}=2 x y(1+2 y), y(0)=-1\)

Short answer

The solution to the initial-value problem is substantially implicit in the form derived; direct algebraic resolution or approximations may better elucidate explicit solutions.

Step-by-step solution

Identify the Differential Equation

The given problem is a first-order differential equation in the form \( y' = 2xy(1+2y) \) with the initial condition \( y(0) = -1 \). This is a separable differential equation, which means we can express it in the form \( f(y) \, dy = g(x) \, dx \).

Separate Variables

To separate variables, rewrite the equation as \( \frac{dy}{y(1+2y)} = 2x \, dx \). This allows us to integrate each side separately.

Integrate Both Sides

Integrate the left side with respect to \( y \) and the right side with respect to \( x \). For the left side, use partial fraction decomposition: \[\frac{1}{y(1+2y)} = \frac{A}{y} + \frac{B}{1+2y}\]Solving for \( A \) and \( B \) gives \( A = 1 \) and \( B = -\frac{1}{2} \). Thus:\[\int \left( \frac{1}{y} - \frac{1}{2(1+2y)} \right) \, dy = \int 2x \, dx\]

Solve Integrals

Integrate the left side:\[\ln |y| - \frac{1}{2} \ln |1+2y| = x^2 + C\]

Simplify and Solve for C

Simplify the equation by combining the logarithms:\[\ln \left| \frac{y}{(1+2y)^{1/2}} \right| = x^2 + C\]Exponentiate both sides:\[\frac{y}{(1+2y)^{1/2}} = e^{x^2+C}\]Setting \( e^C = C_1 \), we have:\[ y = C_1 (1+2y)^{1/2} e^{x^2}\]

Apply Initial Condition

Use the initial condition \( y(0) = -1 \) to find \( C_1 \):\[-1 = C_1 (1 - 2)^{1/2} \rightarrow C_1 = 1.\]Applying the initial condition modifies the equation to:\[ y = (1 + 2y)^{1/2} e^{x^2}\]

Solve the Initial Value Problem

Substitute \( C_1 \) back into the equation, we solve for \( y \) in terms of \( x \). The final solution will have an implicit form involving \( y \) and \( x \). The correct solution entails trial and improvement or direct algebraic manipulations to extract \( y \) in terms of \( x \).

Finalize the Solution

The solution is in the form of a complex algebraic expression involving \( y \) and \( x \). The initial condition suggests the equation balance when evaluating the function at the starting point. However, explicitly expressing \( y \) in an elementary function depends on deeper algebraic manipulation if possible.

Key concepts

Initial Value Problem

An initial value problem involves a differential equation along with an extra condition, known as the initial condition. This condition specifies the value of the unknown function at a specified point, commonly denoted as the initial time. For example, in the given problem, the differential equation is presented as \( y' = 2xy(1+2y) \) with an initial condition \( y(0) = -1 \).
This means when the variable \( x \) is 0, the value of the function \( y \) should be -1.
These problems are vital because they uniquely determine the specific solution to a differential equation from among many possible solutions.
To solve these types of problems, we first need to find the general solution to the differential equation and then apply the initial condition to find any constants involved.
This provides the specific solution that passes through the given point of the initial condition. Understanding the initial condition is crucial as it transforms the problem from an abstract form into a practical one with a clear starting point.

Separation of Variables

Separation of variables is a method used for solving ordinary differential equations, particularly useful when the differential equation can be expressed as a product of a function of \( y \) and a function of \( x \).
This technique is effective because it allows us to rewrite the differential equation in a form where all terms involving \( y \) are on one side and all terms involving \( x \) are on the other.
In the exercise, the equation \( y' = 2xy(1+2y) \) can be rearranged to \( \frac{dy}{y(1+2y)} = 2x \, dx \).
This separation of terms makes it possible for us to integrate both sides separately.
The power of this method lies in breaking the problem into smaller, more manageable parts that can be solved independently through integration. By integrating both sides, we achieve a relationship between \( y \) and \( x \), which helps in progressing towards finding the specific solution.

Partial Fraction Decomposition

Partial fraction decomposition is a crucial algebraic technique used to simplify complex rational expressions for easier integration.
When we have an expression where variables are located in both the numerator and the denominator, it becomes necessary to rewrite it in a form that is easier to integrate. This is where partial fraction decomposition becomes handy.
In our problem, we decomposed \( \frac{1}{y(1+2y)} \) to \( \frac{A}{y} + \frac{B}{1+2y} \).
By solving for the constants \( A = 1 \) and \( B = -\frac{1}{2} \), we were then able to integrate this expression more easily.
This process involves expressing the fraction as a sum of simpler fractions, which then can be integrated individually.
It's a powerful method that not only simplifies integration of rational expressions but also plays a crucial role in solving differential equations like the one in our exercise.