Problem 45 - Calculus and its applications

Question

Solve for \(y .\) \(y^{\prime}=5 y^{-2} ; \quad y=3\) when \(x=2\)

Step-by-Step Solution

Answer

The solution is \( y = (15x - 3)^{1/3} \).

Step 1: Identify the Differential Equation Type

The given differential equation is \( y' = 5y^{-2} \). This is a separable differential equation because it can be rewritten in the form \( N(y) \, dy = M(x) \, dx \).

Step 2: Rearrange the Equation

Rearrange the equation to separate variables: \[y^2 \, dy = 5 \, dx\]This allows us to integrate both sides with respect to their respective variables.

Step 3: Integrate Both Sides

Integrate the left side with respect to \( y \) and the right side with respect to \( x \):\[\int y^2 \, dy = \int 5 \, dx\]This yields:\[\frac{y^3}{3} = 5x + C\]where \( C \) is the constant of integration.

Step 4: Solve for the Constant of Integration

Use the initial condition \( y = 3 \) when \( x = 2 \) to find \( C \):\[\frac{3^3}{3} = 5(2) + C\]Simplifying this:\[9 = 10 + C \]Therefore, \( C = -1 \).

Step 5: General Solution and Final Answer

Substitute the value of \( C \) back into the integrated equation:\[y^3 = 15x - 3\]Solve for \( y \):\[y = (15x - 3)^{1/3}\]This is the solution for \( y \) in terms of \( x \).

Key Concepts

Differential Equation

A differential equation is an equation that involves the derivatives of a function. In simpler terms, it's a mathematical statement expressing how a quantity changes as it relates to other quantities. In our exercise, we are dealing with a first-order differential equation, which involves the first derivative of the function.

• Notation: A common notation for the first derivative of a function, often denoted as \( y' \) or \( \frac{dy}{dx} \), indicates the rate of change of \( y \) with respect to \( x \).
• Types: Differential equations can be classified into various types, such as ordinary differential equations (ODEs), which involve derivatives with respect to one independent variable, and partial differential equations (PDEs), which involve derivatives with respect to multiple variables.

Separable Differential EquationsA crucial aspect of our problem is recognizing it as a separable differential equation. This type allows us to separate variables, meaning we can rewrite the equation so that all terms involving \( y \) are on one side, and all terms involving \( x \) are on the other side.

Integration

Integration is the process of finding the original function from its derivative, and it is a key tool for solving differential equations. When dealing with separable differential equations, like the one in our exercise, integration plays a crucial role in finding the function that satisfies the equation.

• Process: To integrate each side of the separated equation, we find the antiderivative, which is essentially the reverse of differentiation.
• Constant of Integration: When we integrate, we must include a constant of integration \( C \) because the process of differentiation doesn't include that constant. This constant accounts for all possible vertical shifts of the antiderivative function.

In our case, integrating the left side \( \int y^2 \, dy \) and the right side \( \int 5 \, dx \) leads us to the equation \( \frac{y^3}{3} = 5x + C \), where \( C \) is the integration constant.

Initial Conditions

Initial conditions are specific values given in a differential equation problem that allow us to find the particular solution among the family of solutions. These conditions describe the behavior of the solution at a specific point, and they are necessary for solving problems exactly.

• Utility: Initial conditions are useful because they help us determine the exact value of the constant of integration \( C \) from our previous integration process.
• Example: In our example, we are given the initial condition \( y = 3 \) when \( x = 2 \). This means that when we plug these values into the integrated equation, \( \frac{y^3}{3} = 5x + C \), we can solve for \( C \).

By substituting these specific values, we find \( C = -1 \), giving us the precise solution: \( y = (15x - 3)^{1/3} \). This tailored solution reflects the unique curve that matches the given initial condition, distinguishing it from any other potential solutions that might exist without such specific constraints.