What is the solution for y^{''}+16y=0,y(0)=cos(4x) ?

The solution for y^{''}+16y=0,y(0)=cos(4x) is y=c_{1}cos(4x)+c_{2}sin(4x)

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y^{''}+16y=0,y(0)=cos(4x)

y^{^{''}} + 16 y = 0, y (0) = cos (4 x)

y = c_{1} cos (4 x) + c_{2} sin (4 x)

Solution steps

y^{''} (x) + 16 y = 0

Solve linear ODE:

y = c_{1} cos (4 x) + c_{2} sin (4 x)

y = c_{1} cos (4 x) + c_{2} sin (4 x)

y^{^{''}} - y^{^{'}} - y = 0, y^{1} (0) = e^{x}, y^{2} (0) = x

12 y^{^{''}} - 5 y^{^{'}} - 3 y = 0

y^{^{''}} + \frac{49}{\frac{4}{32}} y = 0

y^{^{''}} + 9 y = 0, y^{^{'}} (3 π) = 9, y (0) = 9

(y^{^{''}}) + 9 y = 0

What is the solution for y^{''}+16y=0,y(0)=cos(4x) ?
The solution for y^{''}+16y=0,y(0)=cos(4x) is y=c_{1}cos(4x)+c_{2}sin(4x)