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

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

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

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

y = c_{1} cos (2 t) + c_{2} sin (2 t)

Solution steps

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

Solve linear ODE:

y = c_{1} cos (2 t) + c_{2} sin (2 t)

y = c_{1} cos (2 t) + c_{2} sin (2 t)

\frac{d ^{2} y}{d x ^{2}} - 6 \frac{d y}{d x} + 3 y = 0

\frac{d ^{4} y}{d x ^{4}} + 6 \frac{d ^{2} y}{d x ^{2}} + 8 y = 0

y^{^{''}} + 9 y^{^{'}} - 20 y = 0

y^{^{''}} - 20 y^{^{'}} + 64 y = 0

f^{^{''}} (t) - c f (t) = 0

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