Solve The Separable Differential Equation Subject To The Initial Condition

Solving Separable Differential Equations with Initial Conditions

Separable differential equations are a fundamental type of differential equation that can be solved using integration techniques. Understanding how to solve these equations, particularly when an initial condition is provided, is crucial in various fields, from physics and engineering to economics and biology. This comprehensive guide will delve into the process, providing examples and explanations to solidify your understanding.

What are Separable Differential Equations?

A separable differential equation is a first-order differential equation that can be written in the form:

dy/dx = f(x)g(y)

where f(x) is a function of x alone, and g(y) is a function of y alone. The key characteristic is that the variables x and y can be separated onto opposite sides of the equation. This allows us to integrate both sides independently, leading to a solution.

The Solution Process: A Step-by-Step Guide

Solving a separable differential equation involves several key steps:

Step 1: Separate the Variables

Rewrite the equation so that all terms involving y (and dy) are on one side, and all terms involving x (and dx) are on the other. This usually involves algebraic manipulation. For example, if you have:

dy/dx = x²y

You would separate the variables as follows:

(1/y) dy = x² dx

Step 2: Integrate Both Sides

Once the variables are separated, integrate both sides of the equation with respect to their respective variables. Remember to include the constant of integration, typically denoted as 'C'.

∫(1/y) dy = ∫x² dx

This yields:

ln|y| = (x³/3) + C

Step 3: Solve for y (if possible)

This step involves manipulating the equation algebraically to express y explicitly as a function of x. This might involve exponentiation, logarithms, or other algebraic techniques. In our example:

ln|y| = (x³/3) + C

Taking the exponential of both sides:

|y| = e^((x³/3) + C) = e^(x³/3) * e^C

Since e^C is an arbitrary constant, we can replace it with another constant, say 'A':

y = ±Ae^(x³/3)

This can often be simplified further, depending on the specific problem.

Step 4: Apply the Initial Condition

The initial condition is a given value of y at a specific value of x (often y(0) = some value). This condition allows us to determine the specific value of the constant A. Let's assume the initial condition is y(0) = 2:

2 = ±Ae^(0³/3) = ±A

Therefore, A = 2 (we'll choose the positive solution for simplicity). Our final solution becomes:

y = 2e^(x³/3)

Examples: Illustrating the Process

Let's work through a few more examples to solidify our understanding.

Example 1:

Solve the differential equation dy/dx = 2xy, subject to the initial condition y(0) = 1.

1. Separate variables: (1/y) dy = 2x dx

2. Integrate: ∫(1/y) dy = ∫2x dx => ln|y| = x² + C

3. Solve for y: |y| = e^(x² + C) = Ae^(x²)

4. Apply initial condition: 1 = Ae^(0²) => A = 1

Solution: y = e^(x²)

Example 2:

Solve the differential equation dy/dx = (1+y²)/(1+x²), subject to the initial condition y(0) = 1.

1. Separate variables: (1/(1+y²)) dy = (1/(1+x²)) dx

2. Integrate: ∫(1/(1+y²)) dy = ∫(1/(1+x²)) dx => arctan(y) = arctan(x) + C

3. Solve for y: y = tan(arctan(x) + C)

4. Apply initial condition: 1 = tan(arctan(0) + C) => 1 = tan(C) => C = π/4

Solution: y = tan(arctan(x) + π/4)

Example 3: A slightly more complex scenario

Solve the differential equation (x²+1)dy/dx = xy², subject to the initial condition y(0) = 1.

1. Separate variables: (1/y²) dy = x/(x²+1) dx

2. Integrate: ∫y⁻² dy = ∫x/(x²+1) dx. Notice that the integral on the right is best approached using substitution. Let u = x² + 1, then du = 2x dx. This gives:

∫y⁻² dy = (1/2)∫(1/u) du

This yields: -1/y = (1/2)ln|x²+1| + C

3. Solve for y: -1/y = (1/2)ln(x²+1) + C => y = -1/((1/2)ln(x²+1) + C)

4. Apply initial condition: 1 = -1/((1/2)ln(1) + C) => 1 = -1/C => C = -1

Solution: y = -1/((1/2)ln(x²+1) - 1)

Handling Different Initial Conditions

The initial condition can appear in various forms, such as y(a) = b, where 'a' and 'b' are constants. The method remains the same; substitute the values of x = a and y = b into the general solution to find the value of the integration constant C.

Common Mistakes to Avoid

• Forgetting the constant of integration: This is a crucial step. Without the constant of integration, your solution will not be general enough and may not satisfy the initial condition.
• Incorrect separation of variables: Ensure you correctly isolate the terms involving x and y. Any mistakes here will propagate through the rest of the solution.
• Errors in integration: Double-check your integration techniques. Incorrect integration will lead to an incorrect solution.
• Algebraic mistakes: Carefully perform the algebraic manipulations involved in solving for y and applying the initial condition.

Conclusion

Solving separable differential equations with initial conditions is a fundamental skill in calculus and its applications. By mastering the step-by-step process outlined in this guide, you'll be equipped to tackle a wide range of problems and gain a deeper understanding of this important class of differential equations. Remember to practice regularly, work through various examples, and pay close attention to detail to avoid common mistakes. With consistent practice, you'll develop confidence and proficiency in solving these equations. This understanding forms a strong base for tackling more complex differential equations in the future.