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

The solution for f^{''}(x)=-f(x) is f(x)=c_{1}cos(x)+c_{2}sin(x)

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

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

f (x) = c_{1} cos (x) + c_{2} sin (x)

Solution steps

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

Solve linear ODE:

f (x) = c_{1} cos (x) + c_{2} sin (x)

f (x) = c_{1} cos (x) + c_{2} sin (x)

y^{^{''}} - n^{2} π^{2} y = 0

y^{^{''}} + 2 y^{^{''}} - 3 y = 0

y^{^{''}} - y = 0, y (0) = \frac{1}{5}, y (1) = \frac{1}{5} cosh (1)

\frac{d ^{2} y}{d x ^{2}} - 2 \frac{d y}{d x} - 3 y = 0

y^{^{''}} - 2 y^{^{'}} - 8 y = 0, y (0) = 1, y^{^{'}} (0) = - 3

What is the solution for f^{''}(x)=-f(x) ?
The solution for f^{''}(x)=-f(x) is f(x)=c_{1}cos(x)+c_{2}sin(x)