Problem 14 - Calculus: Early Transcendentals

Question

Find the solution of the differential equation that satisfies the given initial condition. \( x + 3y^2 \sqrt {x^2 + 1} \frac {dy}{dx} = 0, y(0) = 1 \)

Step-by-Step Solution

Answer

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

Step 1: Separate Variables

Start by rearranging the given differential equation: \( x + 3y^2 \sqrt{x^2 + 1} \frac{dy}{dx} = 0 \). Move \(x\) to the other side: \[ 3y^2 \sqrt{x^2 + 1} \frac{dy}{dx} = -x \]. Then, divide both sides by \(3y^2 \sqrt{x^2 + 1}\): \[ \frac{dy}{dx} = -\frac{x}{3y^2 \sqrt{x^2 + 1}} \]. Next, rearrange to separate variables: \[ 3y^2 dy = -\frac{x dx}{\sqrt{x^2 + 1}} \]. This has the variables separated between two sides.

Step 2: Integrate Both Sides

Integrate both sides of the equation obtained in Step 1. The left side is \( \int 3y^2 \ dy \) which simplifies to \( y^3 \). The right side requires a substitution: let \( u = x^2 + 1 \), then \( du = 2x dx \), giving \(-\int \frac{x dx}{\sqrt{x^2+1}} = -\frac{1}{2}\int \frac{du}{\sqrt{u}} = -\sqrt{u} = -\sqrt{x^2+1}\). This integration step results in \[ y^3 = -\sqrt{x^2+1} + C \].

Step 3: Apply Initial Condition

Use the initial condition \( y(0) = 1 \) to find the constant \( C \). Substitute \( y = 1 \) and \( x = 0 \) into the integrated equation from Step 2: \[ 1^3 = -\sqrt{0^2+1} + C \], which simplifies to \( 1 = -1 + C \). Solving for \( C \) gives \( C = 2 \).

Step 4: Write the General Solution

Now substitute \( C = 2 \) back into the integrated equation: \[ y^3 = -\sqrt{x^2+1} + 2 \]. This represents the solution to the differential equation satisfying the initial condition.

Key Concepts

Initial Conditions

In differential equations, an initial condition is a helpful piece of information that allows us to find a specific solution from a family of general solutions. The solution to a differential equation is not always unique, as many functions can satisfy the equation. Initial conditions, like the given point \( y(0) = 1 \) for this problem, specify exact values that the solution must pass through. This restriction cuts down the number of possible solutions to just one, making it possible to pinpoint the exact curve representing the solution. When we substitute these values into our integrated equation, we can solve for any constants that arise from the integration process, giving us a unique curve that fits the conditions exactly.

Integration Techniques

To solve differential equations, we often use a range of integration techniques. One common method involves integrating both sides of an equation once the variables have been separated. For example, in our solution, we integrated \( \int 3y^2 \, dy \) on the left side, transforming it into \( y^3 \).

Integrating can sometimes be straightforward, as with basic power or polynomial functions. However, it can become more complicated when dealing with radicals or trigonometric functions. Techniques such as substitution or integration by parts may be used to simplify and solve these integrals. In this example, substitution was crucial in solving \( \int \frac{x}{\sqrt{x^2+1}} \, dx \), which was not directly integrable using simple techniques.

Mastering these techniques requires practice, as it involves recognizing which method fits a given integral best and applying it correctly.

Separation of Variables

Separation of variables is a powerful method used to solve differential equations. The idea is to manipulate the equation so that all terms involving one variable appear on one side of the equation, and all terms involving the other variable appear on the opposite side.

For the equation \( x + 3y^2 \sqrt{x^2 + 1} \frac{dy}{dx} = 0 \), we achieve separation by first isolating \( \frac{dy}{dx} \) and rearranging the terms to obtain \( 3y^2 \, dy = -\frac{x}{\sqrt{x^2 + 1}} \, dx \). This transforms our equation into a form where we can integrate each side separately.

This technique allows us to reduce complex problems to more manageable integrals which can be solved individually. By breaking the equation into two integrals, each dependent on a single variable, we progressively step closer to finding a solution. Separation of variables is particularly helpful with first-order differential equations, as seen in our current exercise.

Variable Substitution

Variable substitution is a strategic method used to simplify integrals, especially when faced with complex expressions that are difficult to integrate directly. In our exercise, after separating variables, we needed to integrate \( \int \frac{x}{\sqrt{x^2+1}} \, dx \). Direct integration was not straightforward, so substitution came to the rescue.

By setting \( u = x^2 + 1 \), we were able to simplify the integration process. This change of variables transformed \( \, x dx \) into \( \, \frac{1}{2} du \). Substitution helps to convert difficult integrals into forms that are much easier to tackle.

Using substitutions can transform a tyranny of complex algebra into a manageable problem, saving time and making the integration process simpler. Understanding when and how to apply substitutions is key to mastering differential equation solutions.