What is the solution for y^{''}+4y=0,y(0)=cos(2x) ?

The solution for y^{''}+4y=0,y(0)=cos(2x) is y=c_{1}cos(2x)+c_{2}sin(2x)

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

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

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

Solution steps

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

Solve linear ODE:

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

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

(y^{^{''}} + 2 y^{^{'}} + 5 y)^{3} = 0

5 y^{^{''}} + 3 y^{^{'}} + 9 y = 0

y^{^{''}} - 22 y^{^{'}} + 2 y = 0

1 y^{^{''}} + 10 y^{^{'}} + \frac{32}{10} y = 0

y^{^{''}} - 20 y^{^{'}} - a y = 0, y (0) = 1, y^{^{'}} (0) = 15

What is the solution for y^{''}+4y=0,y(0)=cos(2x) ?
The solution for y^{''}+4y=0,y(0)=cos(2x) is y=c_{1}cos(2x)+c_{2}sin(2x)