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

The solution for y^{''}+4y=cos(4x),y(0)=0,y^'(0)=0 is y= 1/12 cos(2x)-1/12 cos(4x)

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

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

y = \frac{1}{12} cos (2 x) - \frac{1}{12} cos (4 x)

Solution steps

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

Solve linear ODE:

y = \frac{1}{12} cos (2 x) - \frac{1}{12} cos (4 x)

y = \frac{1}{12} cos (2 x) - \frac{1}{12} cos (4 x)

\frac{\partial}{\partial x} (13 (3^{x}))

\frac{d}{d x} (\frac{x}{30})

\int \frac{9 x ^{2} - 3 x + 19}{( x ^{2} + 4 ) ( x - 3 )} d x

\int_{- π}^{π} x sin (x) d x

3 (x^{3} - x), 0, - 1, 1

What is the solution for y^{''}+4y=cos(4x),y(0)=0,y^'(0)=0 ?
The solution for y^{''}+4y=cos(4x),y(0)=0,y^'(0)=0 is y= 1/12 cos(2x)-1/12 cos(4x)