Find the solution of the equation y' + 2y = 0.

Practice Questions

Q1
Find the solution of the equation y' + 2y = 0.
  1. y = Ce^(-2x)
  2. y = Ce^(2x)
  3. y = 2Ce^x
  4. y = Ce^x

Questions & Step-by-Step Solutions

Find the solution of the equation y' + 2y = 0.
  • Step 1: Identify the equation. The equation is y' + 2y = 0.
  • Step 2: Recognize that this is a first-order linear differential equation.
  • Step 3: Rewrite the equation in standard form: y' = -2y.
  • Step 4: Separate the variables by dividing both sides by y: dy/y = -2 dx.
  • Step 5: Integrate both sides. The left side becomes ln|y| and the right side becomes -2x + C (where C is the constant of integration).
  • Step 6: Write the result of the integration: ln|y| = -2x + C.
  • Step 7: Exponentiate both sides to solve for y: |y| = e^(-2x + C).
  • Step 8: Simplify the right side: |y| = e^C * e^(-2x).
  • Step 9: Let C' = e^C (a new constant), so |y| = C' * e^(-2x).
  • Step 10: Remove the absolute value by allowing C' to be any real number (positive or negative): y = Ce^(-2x), where C is a constant.
  • First-Order Linear Differential Equations – These equations involve the first derivative of a function and can be solved using integrating factors or separation of variables.
  • Homogeneous Equations – The given equation is homogeneous, meaning it can be expressed in the form y' + P(x)y = 0, which simplifies the solution process.
  • Exponential Solutions – The general solution to such equations often involves exponential functions, specifically in the form y = Ce^(kx).
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