What is the solution to the equation y' + 2y = 0?

Practice Questions

Q1
What is the solution to 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

What is the solution to the equation y' + 2y = 0?
  • Step 1: Identify the equation. We have y' + 2y = 0.
  • Step 2: Recognize that y' means the derivative of y with respect to x.
  • Step 3: Rearrange the equation to isolate y'. This gives us y' = -2y.
  • Step 4: This is a separable equation, meaning we can separate y and x. Rewrite it as dy/dx = -2y.
  • Step 5: Separate the variables by dividing both sides by y and multiplying both sides by dx. This gives us (1/y) dy = -2 dx.
  • Step 6: Integrate both sides. The left side becomes ln|y| and the right side becomes -2x + C, where C is the constant of integration.
  • Step 7: Write the result of the integration: ln|y| = -2x + C.
  • Step 8: Exponentiate both sides to solve for y. This gives us |y| = e^(-2x + C).
  • Step 9: Rewrite e^C as a new constant, which we can call C. So, |y| = Ce^(-2x).
  • Step 10: Since y can be positive or negative, we can drop the absolute value, giving us y = Ce^(-2x).
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