What is the particular solution of dy/dx = 4x with the initial condition y(0) =

₹0.0

Question: What is the particular solution of dy/dx = 4x with the initial condition y(0) = 1?

Options:

Correct Answer: y = 2x^2 + 1

Solution:

Integrating gives y = 2x^2 + C. Using the initial condition, C = 1.

What is the particular solution of dy/dx = 4x with the initial condition y(0) = 1?

What is the particular solution of dy/dx = 4x with the initial condition y(0) = 1?

Step 1: Start with the differential equation dy/dx = 4x.
Step 2: Integrate both sides with respect to x. This means we find the antiderivative of 4x.
Step 3: The integral of 4x is 2x^2 + C, where C is a constant.
Step 4: Now we have the general solution: y = 2x^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) = 2(0)^2 + C = 1.
Step 7: This simplifies to C = 1.
Step 8: Substitute C back into the general solution to get the particular solution: y = 2x^2 + 1.

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