Consider The Differential Equation Dy Dx Xy 2

The differential equation $ \frac{dy}{dx} = x y^2 $ serves as a classic introductory example of a separable first-order ordinary differential equation. And it appears frequently in calculus curricula and mathematical modeling contexts because it demonstrates the fundamental technique of separating variables clearly and yields a solution family that illustrates the concept of singular solutions and domain restrictions. Understanding how to solve this equation provides a crucial foundation for tackling more complex non-linear dynamics encountered in physics, biology, and engineering.

Understanding the Structure of the Equation

Before diving into the solution mechanics, Recognize the anatomy of the problem — this one isn't optional. On the flip side, it is non-linear because the dependent variable $ y $ appears raised to the second power. Still, the equation $ \frac{dy}{dx} = x y^2 $ is classified as first-order because the highest derivative present is the first derivative. Most importantly for our solution strategy, it is separable.

A separable differential equation can be written in the form: $ \frac{dy}{dx} = g(x) h(y) $ In this specific case, $ g(x) = x $ and $ h(y) = y^2 $. The defining characteristic of a separable equation is that the variables $ x $ and $ y $ can be algebraically manipulated to opposite sides of the equality, allowing for direct integration with respect to their respective variables. This property transforms a differential problem into an integral calculus problem And that's really what it comes down to..

Step-by-Step Solution Process

Solving $ \frac{dy}{dx} = x y^2 $ follows a rigid, algorithmic procedure. Mastering these steps ensures accuracy when initial conditions are introduced or when the algebra becomes more complicated The details matter here..

1. Separate the Variables

The first step involves treating the derivative $ \frac{dy}{dx} $ as a ratio of differentials $ dy $ and $ dx $. While this is technically an abuse of notation (derivatives are limits, not fractions), it is a universally accepted mnemonic device in introductory differential equations.

We divide both sides by $ y^2 $ (assuming $ y \neq 0 $) and multiply both sides by $ dx $: $ \frac{1}{y^2} , dy = x , dx $ Rewriting $ \frac{1}{y^2} $ as $ y^{-2} $ often makes the integration step more intuitive: $ y^{-2} , dy = x , dx $

2. Integrate Both Sides

Now we apply the integral operator to both sides. Remember to include the constant of integration. It is standard practice to add a single arbitrary constant $ C $ to one side (usually the $ x $-side) rather than adding constants to both sides, as the difference of two arbitrary constants is just another arbitrary constant Practical, not theoretical..

$ \int y^{-2} , dy = \int x , dx $

Applying the power rule for integration ($ \int u^n du = \frac{u^{n+1}}{n+1} $ for $ n \neq -1 $):

• Left side: $ \int y^{-2} dy = \frac{y^{-1}}{-1} = -y^{-1} = -\frac{1}{y} $
• Right side: $ \int x dx = \frac{x^2}{2} $

Combining these results: $ -\frac{1}{y} = \frac{x^2}{2} + C $

3. Solve for $ y $ (Explicit Solution)

The result above is an implicit solution. Usually, we prefer an explicit solution where $ y $ is isolated as a function of $ x $.

Multiply the equation by $ -1 $: $ \frac{1}{y} = -\frac{x^2}{2} - C $

Since $ C $ is an arbitrary constant, $ -C $ is also an arbitrary constant. We can rename $ -C $ as a new constant $ K $ (or simply keep it as $ C $, understanding it has flipped sign). Let's use $ C $ for simplicity, acknowledging the sign change: $ \frac{1}{y} = -\frac{x^2}{2} + C $

To combine terms on the right, find a common denominator (2): $ \frac{1}{y} = \frac{-x^2 + 2C}{2} $

Invert both sides to solve for $ y $: $ y = \frac{2}{2C - x^2} $

Again, since $ 2C $ is just an arbitrary constant, we can rename it $ K $ (or $ C $ again). The general explicit solution is typically written as: $ y = \frac{2}{C - x^2} \quad \text{or} \quad y = \frac{-2}{x^2 + C} $ Both forms are equivalent; the constant $ C $ simply absorbs the sign differences.

This is where a lot of people lose the thread Most people skip this — try not to..

The Critical Case: The Singular Solution

A common pitfall for students occurs in Step 1, where we divided by $ y^2 $. Division by a variable quantity is only valid if that quantity is non-zero. By assuming $ y \neq 0 $, we implicitly lost a potential solution.

We must check if $ y = 0 $ satisfies the original differential equation. Substitute $ y = 0 $ into $ \frac{dy}{dx} = x y^2 $:

• Left Hand Side (LHS): $ \frac{d(0)}{dx} = 0 $
• Right Hand Side (RHS): $ x (0)^2 = 0 $

Since $ 0 = 0 $, $ y = 0 $ is a valid solution Worth knowing..

This solution, $ y = 0 $, is called a singular solution (or sometimes an equilibrium solution). It cannot be obtained from the general solution $ y = \frac{2}{C - x^2} $ for any finite value of the constant $ C $. It exists as an envelope to the family of curves defined by the general solution. In physical models, this often represents a steady state or extinction equilibrium.

Applying Initial Conditions: Initial Value Problems (IVPs)

The general solution represents an infinite family of curves (one for each value of $ C $). To find a particular solution, we need an initial condition, typically given as $ y(x_0) = y_0 $. This pins down the specific value of the arbitrary constant $ C $ That alone is useful..

And yeah — that's actually more nuanced than it sounds.

Example 1: Standard Initial Condition

Solve the IVP: $ \frac{dy}{dx} = x y^2 $, with $ y(0) = 1 $.

1. Use the general solution: $ y = \frac{2}{C - x^2} $.
2. Substitute $ x = 0 $ and $ y = 1 $: $ 1 = \frac{2}{C - 0^2} \implies 1 = \frac{2}{C} \implies C = 2 $
3. Particular solution: $ y = \frac{2}{2 - x^2} $

Domain Consideration: The solution is undefined when the denominator is zero ($ x = \pm \sqrt{2} $). Since the initial condition is at $ x=0 $, the interval of validity is the largest open interval containing $ x=0 $ where the solution is defined and differentiable: $ (-\sqrt{2}, \sqrt{2}) $.

Example 2: Initial Condition Leading to Singular Solution

Solve the IVP: $ \frac{dy}{dx} = x y^2 $, with $ y(0) = 0 $.

1. Attempting the general formula $ y = \frac{2}{C - x^2} $: $ 0 = \frac{2}{C} $ This has no solution for finite $ C $.
2. Recall the singular solution check: $ y = 0 $ satisfies the DE and the initial condition $ y(0)=0 $.
3. Particular solution