Question

In Exercises \(75-78\), solve the differential equation. \(\frac{d y}{d x}=\frac{1}{\sqrt{80+8 x-16 x^{2}}}\)

Short answer

The solved differential equation is \(y=\frac{1}{4} \sin ^{-1}\left(\frac{x-5}{5}\right)+C\)

Step-by-step solution

Re-write the differential equation

First, the given first order differential equation can be re-written to show the separation of variables as \(d y=\frac{1}{\sqrt{80+8 x-16 x^{2}}} d x\)

Manipulate equation to standard form

We need to manipulate and simplify the equation to a format where a standard integral can be identified. The equation can be rewritten as \(d y=\frac{1}{\sqrt{16(5+x)(5-x)}} d x\)

Solve the Integral

By recognizing the integral as a standard form, we can solve it: \(\int d y=\int \frac{1}{4\sqrt{(5+x)(5-x)}} d x\). The standard integral form is \(\int \frac{1}{\sqrt{a^{2}-x^{2}}} d x= \sin ^{-1}\left(\frac{x}{a}\right)+C\), so solved integral is \(y=\frac{1}{4} \sin ^{-1}\left(\frac{x-5}{5}\right)+C\) where C is the constant of integration.

Key concepts

Separation of Variables

The method of separation of variables is a fundamental technique to solve ordinary differential equations. It is particularly useful when the variables involved can be separated so each lies on a different side of the equation. Consider the differential equation provided in the exercise:
\[ \frac{d y}{d x} = \frac{1}{\sqrt{80+8 x-16 x^{2}}} \].
The goal is to arrange the equation so that all terms involving the variable \( y \) are on one side, and all terms involving \( x \) are on the other. This can be seen in the first step of the solution. By rewriting the original equation as \[ d y = \frac{1}{\sqrt{80+8 x-16 x^{2}}} d x \], we have effectively separated the variables.
Separation of variables allows us to integrate both sides of the equation independently. Each side of the equation is treated as a separate integral, leading to a function of \( y \) equal to a function of \( x \), often plus a constant of integration.

Integral Calculus

Integral calculus is the study of integrals and their properties. It's the counterpart to differential calculus, which focuses on derivatives and rates of change. In the context of solving differential equations, integral calculus is used to find a function when its rate of change is known. The process involves finding the antiderivative, or integral, which is a function that describes the accumulation of quantities.
In step two of the problem's solution, it's necessary to simplify the differential equation into a more recognizable form to facilitate the integration process. This step often involves algebraic manipulation to reveal a standard integral form. Once this is achieved, as in step three, the integral on both sides of the separated equation can be solved. The integral \[ \int d y \] simply yields \( y \), while the integral on the right side requires knowledge of standard integral forms to solve.

Standard Integral Forms

Standard integral forms are integral expressions that have well-established solutions. They're central to integral calculus, as recognizing these forms allows for the direct application of formulas to find integrals without the need for more complex calculations.
For example, in the exercise given, the right side of the separated differential equation was manipulated to match a standard integral form: \[ \int \frac{1}{\sqrt{a^{2} - x^{2}}} dx \], where \( a \) is a constant. This standard form corresponds to an inverse trigonometric function, specifically: \[ \int \frac{1}{\sqrt{a^{2} - x^{2}}} dx = \sin^{-1}\left(\frac{x}{a}\right) + C \], where \( C \) is the constant of integration.
In the provided solution, the integral \[ \int \frac{1}{4\sqrt{(5+x)(5-x)}} dx \] is solved using this knowledge. Recognizing this integral form and knowing the corresponding solution are essential skills in integral calculus, enabling students to solve a wide variety of integrals encountered in differential equations.