Question

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

Short answer

The solution is \(y = -\frac{2}{x^2 + 2x - 1}\).

Step-by-step solution

Identify the Type of Differential Equation

The given differential equation is \(y' = y^2(x+1)\). This is a separable differential equation because it can be written in the form \( \frac{dy}{dx} = f(y)g(x) \).

Separate Variables

Rewrite the equation in separable form: \( \frac{dy}{y^2} = (x+1) \, dx \). This separates the variables \(y\) and \(x\) on different sides of the equation.

Integrate Both Sides

Integrate both sides: \( \int \frac{dy}{y^2} = \int (x+1) \, dx \). The left side integrates to \(-\frac{1}{y}\), and the right side integrates to \(\frac{x^2}{2} + x + C\), where \(C\) is the constant of integration.

Solve for the Integration Constant

Apply the initial condition \(y(0) = 2\). Substitute \(x = 0\) and \(y = 2\) into \(-\frac{1}{y} = \frac{x^2}{2} + x + C\) to find \(-\frac{1}{2} = 0 + 0 + C\). Thus, \(C = -\frac{1}{2}\).

Solve for \(y\)

Substitute \(C = -\frac{1}{2}\) back into the integrated equation: \(-\frac{1}{y} = \frac{x^2}{2} + x - \frac{1}{2}\). Solve for \(y\) to get \(y = -\frac{2}{x^2 + 2x - 1}\).

Key concepts

Initial Value Problem

An initial value problem is a type of differential equation along with a condition that specifies the value of the solution at a particular point. This helps in determining a unique solution to the differential equation. In our example, we have the differential equation \( y' = y^2(x+1) \) with the initial condition \( y(0) = 2 \). The initial condition means that when \( x = 0 \), \( y \) should equal 2. This condition guides us to find the constant term after integration, ensuring the solution fits perfectly through the specified point in the function's graph.

Integration

Integration is the mathematical process used to find a function based on its derivative. It is the reverse operation of differentiation. In separable differential equations like the one we are considering, after we separate the variables, we need to integrate both sides.

• On the left side, we integrate \( \int \frac{dy}{y^2} \), which results in \( -\frac{1}{y} \).
• On the right side, we integrate \( \int (x+1) \, dx \), resulting in \( \frac{x^2}{2} + x + C \), where \( C \) is the constant of integration.

After integrating, the constant \( C \) is defined using the initial condition \( y(0) = 2 \). This ensures that the general solution becomes a specific one, tailored to the problem's constraints.

Differential Equations

Differential equations are mathematical equations involving derivatives, which represent rates of change. They are used to model various phenomena in science and engineering, from population dynamics to heat transfer.
The given equation, \( y' = y^2(x+1) \), requires us to find the function \( y \) whose rate of change is dependent on both \( y \) and \( x \). Solving differential equations often involves techniques like separation of variables, applying initial conditions, and integration to determine the unknown function.

Variable Separation

Variable separation is a technique used to simplify differential equations. We manipulate the equation to get all terms involving \( y \) on one side and all terms involving \( x \) on the other. This method is applicable when the equation can be written in the form \( \frac{dy}{dx} = f(y)g(x) \).
For the equation \( y' = y^2(x+1) \), we rearrange it to \( \frac{dy}{y^2} = (x+1) \, dx \). By isolating the terms, we make it possible to integrate both sides separately, leading us towards a solution. Separating variables not only simplifies the equation but also aids in solving the initial value problem by allowing proper application of integration techniques.