What is the solution for y^{''}+4y=cos(x),y(0)=6,y^'(0)=-6 ?

The solution for y^{''}+4y=cos(x),y(0)=6,y^'(0)=-6 is y= 17/3 cos(2x)-3sin(2x)+1/3 cos(x)

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y^{''}+4y=cos(x),y(0)=6,y^'(0)=-6

y^{^{''}} + 4 y = cos (x), y (0) = 6, y^{^{'}} (0) = - 6

y = \frac{17}{3} cos (2 x) - 3 sin (2 x) + \frac{1}{3} cos (x)

Solution steps

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

Solve linear ODE:

y = \frac{17}{3} cos (2 x) - 3 sin (2 x) + \frac{1}{3} cos (x)

y = \frac{17}{3} cos (2 x) - 3 sin (2 x) + \frac{1}{3} cos (x)

y^{^{''}} - 2 y^{^{'}} + y = \frac{e ^{x}}{2 x}

y^{^{''}} - 2 y^{^{'}} - 3 y = 2 e^{x} - 10 sin (x)

y^{^{''}} - y^{^{'}} - 6 y = 5 t

y^{^{'''}} - y^{^{'}} = 3 t

y^{^{'''}} - y^{^{'}} = 5 t

What is the solution for y^{''}+4y=cos(x),y(0)=6,y^'(0)=-6 ?
The solution for y^{''}+4y=cos(x),y(0)=6,y^'(0)=-6 is y= 17/3 cos(2x)-3sin(2x)+1/3 cos(x)