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

Practice Questions

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

Questions & Step-by-Step Solutions

Find the particular solution of dy/dx = 2x with the initial condition y(0) = 1.
Correct Answer: y = x^2 + 1
  • Step 1: Start with the differential equation dy/dx = 2x.
  • Step 2: Integrate both sides with respect to x. This means we find the integral of 2x.
  • Step 3: The integral of 2x is x^2 + C, where C is a constant.
  • Step 4: Now we have the general solution: y = x^2 + C.
  • Step 5: Use the initial condition y(0) = 1 to find the value of C.
  • Step 6: Substitute x = 0 into the equation: y(0) = 0^2 + C = C.
  • Step 7: Set C equal to 1 because y(0) = 1.
  • Step 8: Now we have the particular solution: y = x^2 + 1.
  • Differential Equations – The question tests the ability to solve a first-order ordinary differential equation using integration.
  • Initial Conditions – The question assesses the understanding of how to apply initial conditions to find the particular solution.
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