Problem 10 - Calculus Single Variable

Question

Solve the given differential equation. $$ \left(4+y^{2}\right) \frac{d y}{d x}=x^{2} $$

Step-by-Step Solution

Answer

The solution is \(y = 2\tan\left(\frac{2x^3}{3} + 2C\right)\).

Step 1: Separate Variables

Start by rewriting the equation to separate variables. Given the equation \((4 + y^2) \frac{dy}{dx} = x^2\), we aim to isolate \(dy\) with terms involving \(y\) and \(dx\) with terms involving \(x\). Divide both sides by \(4 + y^2\) and multiply by \(dx\), yielding \(\frac{dy}{4+y^2} = \frac{x^2}{1} dx\).

Step 2: Integrate Both Sides

Now integrate both sides of the separated equation. Integrate \(\int \frac{dy}{4+y^2}\) on the left and \(\int x^2 dx\) on the right. These integrations yield \(\int \frac{dy}{4+y^2} = \frac{1}{2} \tan^{-1}\left(\frac{y}{2}\right) + C_1\) and \(\int x^2 dx = \frac{x^3}{3} + C_2\).

Step 3: Apply Integration Constants

Since both sides were integrated separately, we combine the constants into a single constant: \(C = C_2 - C_1\). The general solution becomes \(\frac{1}{2} \tan^{-1}\left(\frac{y}{2}\right) = \frac{x^3}{3} + C\).

Step 4: Solve for \(y\)

To explicitly solve for \(y\), manipulate the equation \(\frac{1}{2} \tan^{-1}\left(\frac{y}{2}\right) = \frac{x^3}{3} + C\) and solve for \(y\). Multiply both sides by 2: \(\tan^{-1}\left(\frac{y}{2}\right) = \frac{2x^3}{3} + 2C\). Taking the tangent of both sides gives \(\frac{y}{2} = \tan\left(\frac{2x^3}{3} + 2C\right)\), thus \(y = 2\tan\left(\frac{2x^3}{3} + 2C\right)\).

Key Concepts

Separation of Variables

Separation of variables is a fundamental technique used to solve differential equations, particularly those that are first-order and homogeneous. The main goal is to rearrange the equation so each variable is on a different side. This allows us to integrate each side independently.

In our differential equation \((4 + y^2) \frac{dy}{dx} = x^2\), we start by moving all terms involving \(y\) to one side and all terms involving \(x\) to the other. We do this by dividing both sides by \((4 + y^2)\) and multiplying by \(dx\). The result is \(\frac{dy}{4+y^2} = x^2 \, dx\).

By isolating variables like this, each side of the equation is now ready for integration, which is the next logical step in solving the equation. This technique is quite useful and appears often when dealing with separable differential equations.

Integration

Integration is the process of finding the antiderivative of a function, which is essentially the reverse of differentiation. Once we have separated the variables in a differential equation, integration helps us find solutions that describe the behavior of the function.

For the equation \(\frac{dy}{4+y^2} = x^2 \, dx\), we need to integrate both sides. The left side, \(\int \frac{dy}{4+y^2}\), is a standard integral that results in \(\frac{1}{2} \tan^{-1}\left(\frac{y}{2}\right) + C_1\). The right side is \(\int x^2 \, dx\), which gives us \(\frac{x^3}{3} + C_2\).

• The constant of integration \(C_1\) results from integrating the left side.
• The constant \(C_2\) is from the right side's integration.

Combining the constants into a single constant, \(C = C_2 - C_1\), simplifies the equation and leads us to the general solution of the differential equation.

Trigonometric Substitution

Trigonometric substitution is a clever technique used to integrate certain types of functions, particularly where square roots or squares appear with a variable. Often seen in integration problems, it involves substituting a trigonometric identity to simplify the integral.

In the problem given, the integral \(\int \frac{dy}{4+y^2}\) is solved using a trigonometric substitution. Here, what we recognize is a form that resembles the derivative of \(\tan^{-1}(x)\). Thus, it integrates to \(\frac{1}{2} \tan^{-1}\left(\frac{y}{2}\right)\).

This use of trigonometric identities helps simplify complex expressions and is exceptionally efficient in transforming and solving differential equations involving squares.

Initial Conditions

Initial conditions are an essential part of solving differential equations, as they allow us to find a specific solution from the general solution by determining the constant of integration.

Though initial conditions were not explicitly stated in this problem, they play a crucial role when provided. After integrating and reaching a general solution, you substitute the initial conditions into the solution to solve for the integration constant.

For instance, if an initial condition like \(y(0) = y_0\) were given, you would substitute \(x = 0\) and \(y = y_0\) into the equation \(\frac{1}{2} \tan^{-1}\left(\frac{y}{2}\right) = \frac{x^3}{3} + C\). This allows you to calculate the exact value of \(C\) for the specific solution.

• Initial conditions tailor the general solution to fit specific scenarios.
• They ensure the solution accurately represents the real-world situation or particular problem set.

Utilizing initial conditions ensures that the solution is not just theoretically correct, but also practically applicable.