How do you solve the differential equation given #f''(x)=x^2#, f'(0)=6, f(0)=3?

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Answer 1 · 2017-07-23T05:53:36

#f(x) = 1/12x^4 + 6x + 3#

We just integrate twice.

#int(f''(x)) = int x^2 dx#

#f'(x) = 1/3x^3 + C#

We now solve for #C#.

#6 = 1/3(0)^3 + C#

#C = 6#

Therefore,

#f'(x) = 1/3x^3 + 6#

Accordingly,

#int f'(x) = int 1/3x^3 + 6 dx#

#f(x) = 1/12x^4 + 6x + C#

We solve for #C# one more time.

#3 = 1/12(0)^4 + 6(0) + C#

#3 = C#

The differential equation therefore has solution #f(x) = 1/12x^4 + 6x + 3#.

Hopefully this helps!