Problem 13 - Calculus Single Variable

Question

Solve the given differential equation. $$ \frac{d y}{d x}-3 x^{2} y=0 $$

Step-by-Step Solution

Answer

The solution is \( y = C_1 e^{x^3} \).

Step 1: Recognize the Type of Differential Equation

The given differential equation \( \frac{d y}{d x} - 3x^2 y = 0 \) is a **first-order linear homogeneous differential equation**.

Step 2: Rewrite the Equation

The standard form of a first-order linear differential equation is \( \frac{d y}{d x} + P(x)y = 0 \). Here, \( P(x) = -3x^2 \). The equation is already in the standard form.

Step 3: Identify Solution Method

Since this is a homogeneous equation, it can be solved using the method of separation of variables.

Step 4: Separate the Variables

Rewrite the equation as \( \frac{d y}{y} = 3x^2 \, d x \). Here, all terms involving \( y \) are on one side and terms involving \( x \) are on the other side.

Step 5: Integrate Both Sides

Integrate both sides of the equation: \( \int \frac{1}{y} \, dy = \int 3x^2 \, dx \). This gives \( \ln|y| = x^3 + C \), where \( C \) is the constant of integration.

Step 6: Solve for y

To find \( y \), exponentiate both sides to get \( y = e^{x^3 + C} = e^C \cdot e^{x^3} \). Let \( e^C = C_1 \), where \( C_1 \) is a new constant. Therefore, \( y = C_1 e^{x^3} \).

Key Concepts

First-Order Linear Differential Equations

Differential equations are equations that involve an unknown function and its derivatives. A first-order linear differential equation is a type of differential equation that involves only the first derivative of the unknown function. These equations take the form \( \frac{dy}{dx} + P(x)y = Q(x) \), where \( P(x) \) and \( Q(x) \) are functions of \( x \).

In our exercise, the equation \( \frac{dy}{dx} - 3x^2 y = 0 \) is a perfect example of a first-order linear differential equation. It has only the first derivative of \( y \) with respect to \( x \) and is linear because neither \( y \) nor its derivative is raised to a power other than one. These equations can arise in many real-world phenomena like population growth or decay and electrical circuits.

Homogeneous Differential Equations

A differential equation is called homogeneous if it can be rewritten in the form where there is no constant term (i.e., the right side of the equation is zero). Specifically, for first-order linear equations, it looks like \( \frac{dy}{dx} + P(x)y = 0 \).

In our exercise, the absence of a separate non-zero function on the right-hand side (\( Q(x) = 0 \)) makes \( \frac{dy}{dx} - 3x^2 y = 0 \) a homogeneous differential equation. This type of equation often implies that the solution is proportional to a power of the underlying variable, indicating exponential growth or decay patterns.

Separation of Variables

Separation of variables is a powerful technique for solving differential equations by separating the variables so that each side of the equation depends on only one of the variables.

With our given equation \( \frac{dy}{dx} - 3x^2 y = 0 \), we were able to rewrite it by moving all terms involving \( y \) to one side and terms in \( x \) to the other: \( \frac{dy}{y} = 3x^2 dx \). Now, each side of the equation is independent for integration.

• Step 1: Identify terms with variables \( y \) and \( x \).
• Step 2: Rearrange to isolate differentials, \( \frac{1}{y} dy \) on one side and \( 3x^2 dx \) on the other.

This method makes it easier to integrate each side individually, paving the way to find a solution.

Integration

Integration is the fundamental process of finding a function from its derivative, and it's essential for solving separated differential equations.

After applying separation of variables to \( \frac{dy}{y} = 3x^2 dx \), we integrate both sides:

• The left side becomes \( \int \frac{1}{y} \, dy \), which results in \( \ln|y| \).
• The right side \( \int 3x^2 \, dx \) results in \( x^3 + C \), where \( C \) is the integration constant.

Combining these, we have \( \ln|y| = x^3 + C \).

To solve for \( y \), take the exponential of both sides to yield \( y = e^{x^3 + C} \). Let \( C_1 = e^C \), giving the final form \( y = C_1 e^{x^3} \). This reveals an exponential relationship as expected from a homogeneous linear equation.