Question

For the following problems, find the general solution to the differential equation.\(y^{\prime}=2 t \sqrt{t^{2}+16}\)

Short answer

The solution is \(y(t) = \frac{2}{3}(t^2 + 16)^{3/2} + C.\)

Step-by-step solution

Identify the Type of Differential Equation

The given differential equation is of the form \(y' = 2t \sqrt{t^2 + 16}\). Since it is not explicitly an exact, linear, or separable form, we should look for a function that could be integrated directly.

Recognize the Independent and Dependent Variables

In the equation \(y' = 2t \sqrt{t^2 + 16}\), the independent variable is \(t\) and the dependent variable is \(y\). This recognition helps in determining how to proceed with integration.

Integrate Both Sides

To solve for \(y\), we integrate both sides with respect to \(t\). This gives us:\[ y = \int 2t \sqrt{t^2 + 16} \, dt. \]

Use Substitution to Simplify the Integral

Let \(u = t^2 + 16\), which implies \(du = 2t \, dt\). This substitution simplifies the integral to:\[ y = \int \sqrt{u} \, du. \]

Integrate Using Power Rule

Integrate \(\sqrt{u}\) with respect to \(u\) using the power rule:\[ \int u^{1/2} \, du = \frac{u^{3/2}}{3/2} + C = \frac{2}{3}u^{3/2} + C. \]

Substitute Back to the Original Variable

Replace \(u\) back with \(t^2 + 16\):\[ y = \frac{2}{3}(t^2 + 16)^{3/2} + C. \]

Present the General Solution

The general solution of the differential equation is:\[ y(t) = \frac{2}{3}(t^2 + 16)^{3/2} + C, \]where \(C\) is the arbitrary constant.

Key concepts

Integral Calculus

Integral calculus is a fundamental branch of mathematics that focuses on finding values like areas under curves. It's essential for solving differential equations, like the one given in the problem where we need to find the integral of a function with respect to a variable. This process allows us to determine a function that describes how quantities accumulate over time or space. In this exercise, the differential equation is expressed as a derivative of some unknown function, and our goal is to reverse this by integrating to find that function. Integral calculus helps in evaluating the 'total change' when you've been given a rate of change, such as moving from a derivative back to the original function.

• Helps solve differential equations by finding original functions when given derivatives.
• Used to calculate areas under curves, among other applications.

Understanding integral calculus is key to analyzing and solving physical phenomena described by differential equations.

Substitution Method

The substitution method, a technique in calculus, is crucial when dealing with integrals that are complex or not straightforward to handle directly. In essence, substitution is about changing variables to make an integral easier to evaluate. In our exercise, the expression inside the square root, \( t^2 + 16 \), was replaced by a new variable \( u \). This transformation simplifies the integral's form, turning it into a standard integral that is easier to work with.
The steps usually involve:

• Choosing a part of the integral to substitute with a new variable.
• Expressing the differential \( dt \) in terms of \( du \).
• Rewriting the integral in terms of \( u \) and solving it.
• Substituting back to the original variable to obtain the final result.

By using substitution, we simplified the original problem into a more manageable form, leading us to the solution with less complexity.

Power Rule Integration

The power rule is a basic yet powerful tool in calculus for integrating polynomial expressions. It states that for any expression of the form \( x^n \), the integral is \( \frac{x^{n+1}}{n+1} + C \), provided \( n eq -1 \). This rule applies directly after substitution has simplified the integral in our exercise. Once we reduced the problem to \( \sqrt{u} \) or \( u^{1/2} \), we applied the power rule to integrate.
The following steps highlight the process:

• Convert the square root to an exponent: \( \sqrt{u} = u^{1/2} \).
• Apply the power rule: integrate \( u^{1/2} \) to get \( \frac{u^{3/2}}{3/2} + C \).
• Translate the result back to the original terms using the earlier substitution.

Power rule integration is straightforward but essential for tackling problems involving differential equations, especially when combined with techniques like substitution.