Problem 11 - Calculus Single Variable

Question

Solve the given differential equation. $$ (2+x) \frac{d y}{d x}=y^{2} $$

Step-by-Step Solution

Answer

The general solution is \( y = -\frac{1}{\ln|2+x| + C} \).

Step 1: Separate Variables

First, we rewrite the differential equation to separate the variables. Our goal is to express all terms involving \( y \) on one side and all terms involving \( x \) on the other side. Start with the given equation: \( (2+x) \frac{dy}{dx} = y^2 \). To separate variables, divide both sides by \( y^2 \) and \( (2+x) \), resulting in: \( \frac{1}{y^2} dy = \frac{1}{2+x} dx \).

Step 2: Integrate Both Sides

Now, integrate both sides of the equation with respect to their respective variables. Integrate the left-hand side: \( \int \frac{1}{y^2} dy = \int \frac{1}{(2+x)} dx \). Performing the integrations, we have: \( \int y^{-2} dy = -y^{-1} + C_1 \) and \( \int \frac{1}{2+x} dx = \ln|2+x| + C_2 \).

Step 3: Combine and Simplify

We now equate the expressions obtained from the integration and solve for \( y \). From Step 2, we have: \( -\frac{1}{y} = \ln|2+x| + C \), where \( C = C_2 - C_1 \). Re-arrange to express \( y \) explicitly: \( y = -\frac{1}{\ln|2+x| + C} \).

Step 4: General Solution

The general solution to the differential equation is: \( y = -\frac{1}{\ln|2+x| + C} \), where \( C \) is an arbitrary constant. This expression gives the relationship between \( x \) and \( y \) that satisfies the original differential equation.

Key Concepts

Separation of Variables

Separation of variables is a powerful method used in solving differential equations like the one given in our exercise. The main aim is to separate the variables in such a way that each side of the equation only contains one variable.

In our original equation, \((2+x) \frac{dy}{dx} = y^2\), we need to rearrange it by dividing both sides by \(y^2\) and \(2+x\). This allows us to get all the terms involving \(y\) on one side, and all terms involving \(x\) on the other side. So, it becomes:

• \(\frac{1}{y^2} dy = \frac{1}{2+x} dx\)

This step is critical because it sets up the equation for integration.

When you successfully write the equation this way, you've effectively "separated variables". This method can be used in a variety of situations where it's possible to express the equation in such a form. This step leads us to the next part of the solution, which is integration.

Integration

Integration is the process that allows us to solve the differential equations after we've separated the variables. In our case, once we've reached:

• \(\int \frac{1}{y^2} dy = \int \frac{1}{2+x} dx\)

we can proceed with integrating each side separately.

Integration involves finding the antiderivative of the expressions. On the left-hand side, the integral of \(\frac{1}{y^2}\) with respect to \(y\) is solved as follows:

• \(\int y^{-2} dy = -y^{-1} + C_1\)

For the right-hand side, the integral of \(\frac{1}{2+x}\) with respect to \(x\) is:

• \(\int \frac{1}{2+x} dx = \ln|2+x| + C_2\)

By integrating both sides separately, you capture all possible solutions for \(y\) with respect to \(x\). Please remember that integration often involves an added constant \(C\), representing the family of possible solutions due to the arbitrary nature of antiderivatives. Each constant represents a different specific solution depending on initial or boundary conditions.

General Solution

The general solution to a differential equation encapsulates all possible solutions that satisfy the given equation. After performing integration, we reach a form where we combine both sides of the equation to express the solution.

In our scenario, integrating resulted in:

• \(-\frac{1}{y} = \ln|2+x| + C\)

This equation still includes \(C\), a constant that can take any real value. This constant is vital as it represents the infinite number of solutions that can satisfy our original differential equation.

We re-arrange this equation to solve for \(y\):

• \(y = -\frac{1}{\ln|2+x| + C}\)

This re-formulated equation gives us the general solution. Here, \(C\) remains as an arbitrary constant. The value of \(C\) would be determined by any initial conditions provided, narrowing down to a specific solution. The general solution thus describes a family of curves defined by different values of \(C\), all fulfilling the original equation.