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

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

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y^{''}+4y=4cos(2x),y^'(pi/2)=0,y(0)=1

y^{^{''}} + 4 y = 4 cos (2 x), y^{^{'}} (\frac{π}{2}) = 0, y (0) = 1

y = cos (2 x) - \frac{π}{2} sin (2 x) + x sin (2 x)

Solution steps

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

Solve linear ODE:

y = cos (2 x) - \frac{π}{2} sin (2 x) + x sin (2 x)

y = cos (2 x) - \frac{π}{2} sin (2 x) + x sin (2 x)

x^{^{''}} + 9 x = sin (t)

y^{^{''}} - 2 y^{^{'}} - 3 y = 3 e^{(2 t)}

y^{^{''}} + 2 y^{^{'}} + 10 y = 10

y^{^{''}} + 4 y^{^{'}} + 4 y = e^{x}

y^{^{''}} + 25 y^{^{'}} = - 24 e^{- t}

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