How do you solve the differential equation given #f'(x)=4x#, f(0)=6?

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Answer 1 · 2016-10-31T22:37:15

#y = f(x) = 2x^2 + 6#

Substitute #dy/dx# for #f'(x)#

#dy/dx = 4x#

Use the separation of variable technique:

#dy = 4xdx#

Integrate both sides:

#y = 2x^2 + C#

Evaluate C, using the initial condition:

#6 = 2(0)^2 + C#

#C = 6#