What is the solution for f^{''}(t)+f(t)=0,f^'(0)=1 ?

The solution for f^{''}(t)+f(t)=0,f^'(0)=1 is f(t)=c_{1}cos(t)+sin(t)

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f^{''}(t)+f(t)=0,f^'(0)=1

f^{^{''}} (t) + f (t) = 0, f^{^{'}} (0) = 1

f (t) = c_{1} cos (t) + sin (t)

Solution steps

f^{''} (t) + f (t) = 0

Solve linear ODE:

f (t) = c_{1} cos (t) + sin (t)

f (t) = c_{1} cos (t) + sin (t)

y^{^{''''}} + 5 y^{^{'''}} - 5 y^{^{''}} - 5 y^{^{'}} + \frac{1}{2} y = 0

2 x^{^{''}} + 7 x = 0

(D^{2} + 4 D + 29)^{2} y = 0

y^{^{''}} - 2 y^{^{'}} + 4 y = 0, y (0) = 2, y^{^{'}} (0) = 0

x^{^{''}} + 2 x^{^{'}} + 50 x = 0, x (0) = - \frac{3}{2}, x^{^{'}} (0) = - 2

What is the solution for f^{''}(t)+f(t)=0,f^'(0)=1 ?
The solution for f^{''}(t)+f(t)=0,f^'(0)=1 is f(t)=c_{1}cos(t)+sin(t)