What is the solution for f^{''}(x)=x^2,f^'(0)=6,f(0)=3 ?

The solution for f^{''}(x)=x^2,f^'(0)=6,f(0)=3 is f(x)=(x^4)/(12)+6x+3

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f^{''}(x)=x^2,f^'(0)=6,f(0)=3

f^{^{''}} (x) = x^{2}, f^{^{'}} (0) = 6, f (0) = 3

f (x) = \frac{x ^{4}}{12} + 6 x + 3

Solution steps

f^{''} (x) = x^{2}

Solve

f^{''} (x) = x^{2} : f^{'} (x) = \frac{x ^{3}}{3} + c_{1}

Solve

f^{'} (x) = \frac{x ^{3}}{3} + c_{1} : f (x) = \frac{x ^{4}}{12} + c_{1} x + c_{2}

Apply initial conditions:

f (x) = \frac{x ^{4}}{12} + 6 x + 3

f (x) = \frac{x ^{4}}{12} + 6 x + 3

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What is the solution for f^{''}(x)=x^2,f^'(0)=6,f(0)=3 ?
The solution for f^{''}(x)=x^2,f^'(0)=6,f(0)=3 is f(x)=(x^4)/(12)+6x+3