Math Problem Statement

y(x) = cos(x)?

Solution

Let's revisit the solution process to clarify any confusion. The differential equation we are working with is:

dydx=y2sin(x)\frac{dy}{dx} = y^2 \sin(x)

To solve this, we use separation of variables.

  1. Separate the variables: dyy2=sin(x)dx\frac{dy}{y^2} = \sin(x) \, dx

  2. Integrate both sides: dyy2=sin(x)dx\int \frac{dy}{y^2} = \int \sin(x) \, dx

    The integral on the left side is: y2dy=y1=1y\int y^{-2} \, dy = -y^{-1} = -\frac{1}{y}

    The integral on the right side is: sin(x)dx=cos(x)\int \sin(x) \, dx = -\cos(x)

    So, we have: 1y=cos(x)+C-\frac{1}{y} = -\cos(x) + C

  3. Solve for yy: 1y=cos(x)C\frac{1}{y} = \cos(x) - C y=1cos(x)Cy = \frac{1}{\cos(x) - C}

  4. Apply the initial condition y(0)=1y(0) = 1: 1=1cos(0)C1 = \frac{1}{\cos(0) - C} 1=11C1 = \frac{1}{1 - C} 1C=11 - C = 1 C=0C = 0

    Therefore, the solution to the differential equation with the given initial condition is: y=1cos(x)y = \frac{1}{\cos(x)}

So, the correct solution is not y=cos(x)y = \cos(x), but rather:

y(x)=1cos(x)y(x) = \frac{1}{\cos(x)}

Would you like more details or have any questions?

Here are 5 numbered questions for you to ask next:

  1. Can you explain the method of separation of variables in more detail?
  2. How do you solve initial value problems for differential equations?
  3. What are some common methods for solving first-order differential equations?
  4. Can you show another example of solving a differential equation using separation of variables?
  5. How do you verify the solution of a differential equation?

Tip: When integrating both sides of a differential equation, always check your integration steps carefully to avoid errors.

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Math Problem Analysis

Mathematical Concepts

Differential Equations
Separation of Variables
Integration

Formulas

Separation of variables method
Integral of sine function
Initial value problem solution

Theorems

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Suitable Grade Level

Advanced High School or Undergraduate Level