What is the solution for y^{''}+4y=0,y(0)=4,y^'(0)=1 ?

The solution for y^{''}+4y=0,y(0)=4,y^'(0)=1 is y=4cos(2t)+1/2 sin(2t)

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y^{''}+4y=0,y(0)=4,y^'(0)=1

y^{^{''}} + 4 y = 0, y (0) = 4, y^{^{'}} (0) = 1

y = 4 cos (2 t) + \frac{1}{2} sin (2 t)

Solution steps

y^{''} (t) + 4 y = 0

Solve linear ODE:

y = 4 cos (2 t) + \frac{1}{2} sin (2 t)

y = 4 cos (2 t) + \frac{1}{2} sin (2 t)

\int (cos (9 x)) d x

\frac{d}{d x} (2 x e^{x} + 3 sec (x))

\frac{d}{d x} (ln ((a x)^{n}))

\frac{d y}{d x} = y (1 - y), y (0) = 8

x \to 4 - lim (3 x - 5)

What is the solution for y^{''}+4y=0,y(0)=4,y^'(0)=1 ?
The solution for y^{''}+4y=0,y(0)=4,y^'(0)=1 is y=4cos(2t)+1/2 sin(2t)