What is the solution for f^{''}(x)=-f(x),f(0)=0,f^'(0)=1 ?

The solution for f^{''}(x)=-f(x),f(0)=0,f^'(0)=1 is f(x)=sin(x)

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f^{''}(x)=-f(x),f(0)=0,f^'(0)=1

f^{^{''}} (x) = - f (x), f (0) = 0, f^{^{'}} (0) = 1

f (x) = sin (x)

Solution steps

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

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f (x) = sin (x)

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f (x) = x^{2} + x, at x = - 3

What is the solution for f^{''}(x)=-f(x),f(0)=0,f^'(0)=1 ?
The solution for f^{''}(x)=-f(x),f(0)=0,f^'(0)=1 is f(x)=sin(x)