Question

Find the solution to the initial value problem. $$ y^{\prime}=3 y^{2}(x+\cos x), y(0)=-2 $$

Short answer

The solution is \(y = -\frac{1}{\frac{3x^2}{2} + 3\sin x + \frac{1}{2}}\).

Step-by-step solution

Identify the Type of Differential Equation

The given initial value problem is a first-order ordinary differential equation in the form: \[ y' = 3y^2(x + \cos x) \] This is a separable differential equation because variables can be separated on each side of the equality sign.

Separate Variables

Move terms involving \(y\) to one side and terms involving \(x\) to the other side:\[ \frac{dy}{y^2} = 3(x + \cos x)dx \] This allows us to integrate both sides separately.

Integrate Both Sides

Integrate the left side with respect to \(y\) and the right side with respect to \(x\): \[ \int \frac{dy}{y^2} = \int 3(x + \cos x) dx \] - The integral on the left is a standard integral: \[ \int \frac{dy}{y^2} = -\frac{1}{y} + C_1 \]- The integral on the right can be split and solved: \[ \int 3(x + \cos x) dx = 3\left(\int x \, dx + \int \cos x \, dx \right) = 3\left(\frac{x^2}{2} + \sin x \right) + C_2 \]

Combine the Integrated Results

Combine the results from both sides: \[ -\frac{1}{y} = \frac{3x^2}{2} + 3\sin x + C \] where \(C = C_2 - C_1\) is a constant to be determined using the initial condition.

Apply Initial Condition to Find Constant

Use the initial condition \(y(0) = -2\) to determine \(C\):\[ -\frac{1}{-2} = \frac{3(0)^2}{2} + 3\sin 0 + C \Rightarrow \frac{1}{2} = C \]Therefore, \(C = \frac{1}{2}\).

Write the General Solution

Incorporate the constant \(C\) back into the equation:\[ -\frac{1}{y} = \frac{3x^2}{2} + 3\sin x + \frac{1}{2} \]To solve for \(y\), invert both sides:\[ y = -\frac{1}{\frac{3x^2}{2} + 3\sin x + \frac{1}{2}} \]

Key concepts

First-Order Ordinary Differential Equation

When you come across an equation involving derivatives like \( y' = 3y^2(x + \cos x) \), you're looking at a type of differential equation. Specifically, it's a first-order ordinary differential equation. Let's break this down a bit further.

An ordinary differential equation (often abbreviated as ODE) involves a function and its derivatives. A first-order ODE means it includes only the first derivative of the function. In our case, \( y' \) is the first derivative of \( y \) with respect to \( x \). First-order ODEs are fundamental in modeling various real-world systems, such as the rate of change of populations and the temperature of a cooling object.

When tackling these equations, the goal is typically to find a function \( y(x) \) that satisfies the equation for given initial conditions. Understanding the nature of the equation helps in determining the method to use for the solution.

Separable Differential Equation

A separable differential equation is a special type of first-order ODE. It allows us to rewrite the equation so that all the \( y \) terms are on one side and all the \( x \) terms on the other. This characteristic makes it much easier to solve.In our exercise, the equation \( y' = 3y^2(x + \cos x) \) can be restructured as \( \frac{dy}{y^2} = 3(x + \cos x)dx \). This step is crucial because it transforms the equation into a form that can be integrated separately with respect to \( y \) and \( x \).Here's a quick way to think about separable differential equations:

• Determine if you can split the terms involving \( y \) and \( x \) onto different sides of the equation.
• Integrate both sides individually once the separation is complete.
• After integration, combine the results to find the solution.

Recognizing and separating variables is often the first step toward finding a solution.

Integration

Integration is the process of finding the integral of a function, which is the reverse process of differentiation. It is a vital step in solving differential equations after separating variables.In our problem, once the variables were separated, we performed integration on both sides:

• The left side: \( \int \frac{dy}{y^2} = -\frac{1}{y} + C_1 \).
• The right side: \( \int 3(x + \cos x) dx = 3\left(\frac{x^2}{2} + \sin x\right) + C_2 \).

Each of these integrals finds the "area under the curve" represented by the function, reversing the differentiation process to solve for the original function \( y(x) \). Both indefinite integrals include a constant of integration (\( C_1 \) and \( C_2 \)). When you combine these results, you account for these constants by introducing a single constant \( C \). Understanding how to handle these constants is crucial, particularly when applying initial conditions.

Initial Conditions

Initial conditions provide specific values for the solution at a certain point, often at the beginning of the problem (hence, 'initial'). They allow us to determine the constants of integration from indefinite integrals.In our exercise, the condition \( y(0) = -2 \) was given. This means that when \( x = 0 \), \( y \) must equal \(-2\). By substituting these values into the integrated relationship \( -\frac{1}{y} = \frac{3x^2}{2} + 3\sin x + C \), we can solve for the constant \( C \). Here's how to use initial conditions:

• Substitute the initial condition values into the final equation.
• Solve the equation for the constant of integration \( C \).
• Substitute \( C \) back into the general solution to get the specific solution.

This ensures that the solution satisfies both the differential equation and the initial condition.