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

The solution for f^{''}(x)+f(x)=0,f(0)=1 is f(x)=cos(x)+c_{2}sin(x)

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

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

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

Solution steps

f^{''} (x) + f (x) = 0

Solve linear ODE:

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

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

f^{^{''}} (x) + f (x) = 0, f (0) = 0

\frac{d ^{2} x}{d t ^{2}} - 5 \frac{d x}{d t} + 4 x = 0

\frac{d ^{2} y}{d t ^{2}} + 8 \frac{d y}{d t} + 4 y = 0

20 x^{^{''}} + 9 x^{^{'}} + x = 0, x (0) = 0, x^{^{'}} (0) = 4

y^{^{'''}} + y^{^{''}} - 7 y^{^{'}} + 5 y = 0

What is the solution for f^{''}(x)+f(x)=0,f(0)=1 ?
The solution for f^{''}(x)+f(x)=0,f(0)=1 is f(x)=cos(x)+c_{2}sin(x)