Question

Solve the following equations by separation of the variables: (a) \(y^{\prime}-x y^{3}=0\) (b) \(y^{\prime} \tan ^{-1} x-y\left(1+x^{2}\right)^{-1}=0\); (c) \(x^{2} y^{\prime}+x y^{2}=4 y^{2}\).

Short answer

(a) \(y = \pm \frac{1}{\sqrt{2C + x^2}}\); (b) \(y = Ke^{F(x)}\); (c) \(y = \frac{1}{\frac{4}{x} + \ln|x| + C}\)

Step-by-step solution

Given Equation (a)

The equation is given as \[ y^{\text{prime}} - xy^3 = 0 \]

Separate the Variables

Rewrite the equation to separate the variables:\[ y^{\text{prime}} = xy^3 \]\[ \frac{dy}{dx} = xy^3 \]\[ \frac{dy}{y^3} = x dx \]

Integrate Both Sides

Integrate both sides with respect to their variables:\[ \int y^{-3} dy = \int x dx \]\[ -\frac{1}{2y^2} = \frac{x^2}{2} + C \]where C is the constant of integration.

Solve for y

Rearrange the equation to solve for y:\[ -\frac{1}{2y^2} = \frac{x^2}{2} + C \]\[ y^2 = -\frac{1}{x^2 + 2C} \]\[ y = \pm \frac{1}{\sqrt{2C + x^2}} \]

Given Equation (b)

The equation is given as \[ y^{\text{prime}} \tan^{-1} x - y (1 + x^2)^{-1} = 0 \]

Separate the Variables

Rewrite the equation to separate the variables:\[ y^{\text{prime}} \tan^{-1} x = y (1 + x^2)^{-1} \]\[ \frac{dy}{dx} = \frac{y}{\tan^{-1}x (1 + x^2)} \]\[ \frac{1}{y} dy = \frac{dx}{\tan^{-1}x (1 + x^2)} \]

Integrate Both Sides

Integrate both sides with respect to their variables:\[ \int \frac{1}{y} dy = \int \frac{dx}{\tan^{-1}x (1 + x^2)} \]\[ \ln |y| = \int \frac{dx}{\tan^{-1}x (1 + x^2)} + C \]Determining the right side may require more advanced techniques, so let's denote this as F(x).

Solve for y

Exponentiate both sides to solve for y:\[ y = e^{F(x) + C} \]\[ y = Ke^{F(x)} \]where K is a constant that absorbs the exponent.

Given Equation (c)

The equation is given as \[ x^2 y^{\text{prime}} + xy^2 = 4y^2 \]

Separate the Variables

Rewrite the equation to separate the variables:\[ x^2 \frac{dy}{dx} = 4y^2 - xy^2 \]\[ x^2 \frac{dy}{dx} = y^2 (4 - x) \]\[ \frac{dy}{y^2} = \frac{(4 - x)}{x^2} dx \]

Integrate Both Sides

Integrate both sides with respect to their variables:\[ \int y^{-2} dy = \int \frac{4 - x}{x^2} dx \]\[ -y^{-1} = 4 \int x^{-2} dx - \int x^{-1} dx \]\[ - \frac{1}{y} = -\frac{4}{x} - \ln |x| + C \]where C is the constant of integration.

Solve for y

Rearrange the equation to solve for y:\[ \frac{1}{y} = \frac{4}{x} + \ln |x| + C \]\[ y = \frac{1}{\frac{4}{x} + \ln |x| + C} \]

Key concepts

differential equations

Dealing with differential equations can seem intimidating, but understanding the basics makes it much easier. A differential equation relates a function with its derivatives. Solutions to these equations describe various phenomena such as motion, growth, decay, and more. There are different types of differential equations like ordinary differential equations (ODEs) and partial differential equations (PDEs). In this exercise, we're handling first-order ODEs. They involve the first derivative of the unknown function, which in this case is denoted as y'. First-order ODEs can often be solved using techniques like separation of variables. Let's understand how this technique works.

integration techniques

Integration sits at the heart of solving separable differential equations. Once variables are separated, the equation looks like two integrals that need to be solved. For example, for the equation \(\frac{dy}{y^3} = x \, dx\), we integrate both sides:

• Left side: \(\int y^{-3} dy\)
• Right side: \(\int x \, dx\)

Integrals of power functions, like these, follow straightforward techniques:

• The integral of \(y^{-3} dy\) is \(-\frac{1}{2y^2}\)
• The integral of \(x \, dx\) is \(\frac{x^2}{2}\)

This process transforms a differential equation into an algebraic equation. Specific methods of integration may vary in complexity, but fundamentally it boils down to reversing differentiation. Always remember to add the constant of integration, denoted by \(C\), after performing indefinite integration.

constant of integration

The constant of integration, represented as \(C\), is crucial. When integrating, we’re dealing with indefinite integrals, meaning the result includes an arbitrary constant. This constant accounts for all possible solutions to a differential equation. For instance, after integrating \(\int y^{-3} \, dy\), you get \(-\frac{1}{2y^2} + C\).\(C\) is essential because it allows the formation of a general solution to the differential equation. When solving problems, this constant provides flexibility, representing the infinite family of curves that form solutions. We often determine \(C\) using initial conditions or boundary conditions, which specify the value of the function at a particular point.

initial conditions

Initial conditions specify the value of the function and its derivatives at a certain point, typically to find the particular solution of a differential equation. If an initial condition such as \(y(0) = 1\) is given, it allows you to solve for the constant of integration \(C\). For example, after getting the general solution \(y = \pm \frac{1}{\sqrt{2C + x^2}}\), you substitute the initial condition value into the solution:

• Substitute \(x=0, y=1\)
• Solve for \(C\)

This generates a specific function that satisfies both the differential equation and the initial condition. Initial conditions convert a general solution with a constant of integration into a unique, specific solution.