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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 3 y^{^{'}} + 2 y = e^{3} x cos (e^{x})

y^{^{''}} + 5 y^{^{'}} + y = 4 t^{2} + 8 t + 8

y^{^{''}} + 3 y^{^{'}} = x^{2} sin (x) cos (x)

y^{^{''}} + 2 y^{^{'}} + 5 y = 7 sin (2 t)

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

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