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What is the particular solution of dy/dx = 4x with the initial condition y(0) =
What is the particular solution of dy/dx = 4x with the initial condition y(0) = 1?
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Practice Questions
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Q1
What is the particular solution of dy/dx = 4x with the initial condition y(0) = 1?
y = 2x^2 + 1
y = 4x^2 + 1
y = 2x^2
y = 4x^2 + C
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Integrating gives y = 2x^2 + C. Using the initial condition, C = 1.
Questions & Step-by-step Solutions
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Q
Q: What is the particular solution of dy/dx = 4x with the initial condition y(0) = 1?
Solution:
Integrating gives y = 2x^2 + C. Using the initial condition, C = 1.
Steps: 8
Show Steps
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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