Find the particular solution of dy/dx = 2y with the initial condition y(0) = 1.

Practice Questions

Q1
Find the particular solution of dy/dx = 2y with the initial condition y(0) = 1.
  1. y = e^(2x)
  2. y = e^(2x) + 1
  3. y = 1 + e^(2x)
  4. y = e^(2x) - 1

Questions & Step-by-Step Solutions

Find the particular solution of dy/dx = 2y with the initial condition y(0) = 1.
  • Step 1: Start with the differential equation dy/dx = 2y.
  • Step 2: Recognize that this is a separable equation, meaning we can separate y and x.
  • Step 3: Rewrite the equation as dy/y = 2 dx.
  • Step 4: Integrate both sides. The left side becomes ln|y| and the right side becomes 2x + C (where C is a constant).
  • Step 5: After integrating, we have ln|y| = 2x + C.
  • Step 6: To solve for y, exponentiate both sides to get y = e^(2x + C).
  • Step 7: Rewrite e^(2x + C) as y = e^(2x) * e^C. Let e^C be a new constant, which we can call C (still a constant). So, y = Ce^(2x).
  • Step 8: Now we have the general solution: y = Ce^(2x).
  • Step 9: Use the initial condition y(0) = 1 to find the value of C.
  • Step 10: Substitute x = 0 into the general solution: y(0) = Ce^(2*0) = C.
  • Step 11: Set C equal to 1 because y(0) = 1. So, C = 1.
  • Step 12: Substitute C back into the general solution: y = 1 * e^(2x), which simplifies to y = e^(2x).
  • Separation of Variables – The method used to solve the differential equation by separating the variables y and x.
  • Initial Conditions – Applying the initial condition to find the specific constant in the general solution.
  • Exponential Functions – Understanding the behavior and properties of exponential functions in the context of differential equations.
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