Problem 20 - Differential Equations and Linear Algebra

Question

Find the equation of the curve that passes through the point (-1,1) and whose slope at each point \((x, y)\) is \(x^{2} y^{2}.\)

Step-by-Step Solution

Answer

The equation of the curve that passes through the point (-1,1) and whose slope at each point \((x, y)\) is \(x^{2} y^{2}\) is \(y = \frac{3}{x^3 + 2}\).

Step 1: Write the slope in form of differential equation

We are given that the slope of the curve at each point \((x, y)\) is \(x^{2}y^{2}\). We can write the slope of a curve in terms of derivatives as \(\frac{\mathrm{d}y}{\mathrm{d}x}\). Therefore, we can write the given information in the form of a differential equation as: \[ \frac{\mathrm{d}y}{\mathrm{d}x} = x^2 y^2 \]

Step 2: Solve the differential equation

The given differential equation is separable. We can separate the variables by dividing both sides by \(y^2\) and then integrating both sides with respect to \(x\): \[ \int \frac{1}{y^2} \frac{\mathrm{d}y}{\mathrm{d}x} \, \mathrm{d}x = \int x^2 \, \mathrm{d}x. \] Now, integrate both sides: \[ -\int y^{-2} \, \mathrm{d}y = \int x^2 \, \mathrm{d}x + C. \] After integrating, we get \[ \frac{1}{y} = \frac{1}{3}x^3 + C. \] Now, we need to find the particular solution using the given point \((-1, 1)\).

Step 3: Use the given point to find the particular solution

The given point is \((-1, 1)\). Substitute this point into the equation we found in Step 2: \[ \frac{1}{1} = \frac{1}{3}(-1)^3 + C. \] Solve for \(C\): \[ C = 1 - \frac{1}{3} = \frac{2}{3}. \] Now, substitute \(C\) back into the equation from Step 2: \[ \frac{1}{y} = \frac{1}{3}x^3 + \frac{2}{3}. \] Rearrange the equation to express \(y\) in terms of \(x\): \[ y = \frac{3}{x^3 + 2} \] Hence, the equation of the curve is \[ y = \frac{3}{x^3 + 2}. \]

Key Concepts

Initial Value Problem

An initial value problem is a common task in differential equations where you're asked to find a specific solution to a differential equation when an initial condition is provided. In our exercise, the initial condition is a point through which the curve passes. This point is

• (-1, 1)

We know that the solution curve must pass through this point. The initial value is crucial as it allows us to find a particular solution to a differential equation, distinguishing it from an entire family of solutions. Without the initial value, we would have infinitely many possible solutions.
By substituting our initial condition into the general solution, we can solve for any constants present, thus honing in on the particular solution that satisfies both the differential equation and the initial condition.

Separable Differential Equations

Separable differential equations are a type of differential equation where the variables involved can be separated onto different sides of the equation. This makes them one of the simpler types of differential equations to solve.

In our exercise, the given differential equation is

• \( \frac{\mathrm{d}y}{\mathrm{d}x} = x^2 y^2 \)

The first step in solving a separable differential equation is to algebraically rearrange terms such that each variable and its differential are on opposite sides of the equation. Here, we divide by \( y^2 \) and rearrange, giving

• \( \int \frac{1}{y^2} \, \mathrm{d}y = \int x^2 \, \mathrm{d}x \)

Now, you are left with two integrals that can be evaluated independently. This simplicity is a powerful feature of separable differential equations, making them a popular choice for introductory courses on differential equations.

Integration

Integration is a fundamental concept in calculus used to solve differential equations by finding functions whose derivatives match a given form.

In the context of our exercise, integrating both sides of the separated equation led us to this general solution:

• \( \frac{1}{y} = \frac{1}{3}x^3 + C \)

This equation was obtained by integrating \( -y^{-2} \) with respect to \( y \) and \( x^2 \) with respect to \( x \). Each integral represents the accumulation of area under the curve of the respective functions, resulting in antiderivatives.Once integration is complete, the constant \( C \) represents arbitrary constants of integration arising from the indefinite integrals. This is where the initial condition plays an important role, enabling us to solve for \( C \), thereby providing a specific solution to the initial value problem.
Integration thus serves as a bridge from differential equations to explicit functions, allowing us to examine the behavior of these functions from their rates of change.