What is the solution for y^{''}+y=t-pi/2 ,y(pi/2)=0,y^'(pi/2)=1 ?

The solution for y^{''}+y=t-pi/2 ,y(pi/2)=0,y^'(pi/2)=1 is y=t-pi/2

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y^{''}+y=t-pi/2 ,y(pi/2)=0,y^'(pi/2)=1

y^{^{''}} + y = t - \frac{π}{2}, y (\frac{π}{2}) = 0, y^{^{'}} (\frac{π}{2}) = 1

y = t - \frac{π}{2}

Solution steps

y^{''} (t) + y = t - \frac{π}{2}

Solve linear ODE:

y = t - \frac{π}{2}

y = t - \frac{π}{2}

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

y^{^{''}} - 2 y^{^{'}} - 3 y = 2

y^{^{''}} + 6 y^{^{'}} + 9 y = \frac{e ^{- 3 x}}{( x ^{2} + 1 )}

y^{^{''}} - 2 y^{^{'}} - 3 y = - 4 e^{x} + 3

y^{^{''}} + y = sin (t) + cos (t)

What is the solution for y^{''}+y=t-pi/2 ,y(pi/2)=0,y^'(pi/2)=1 ?
The solution for y^{''}+y=t-pi/2 ,y(pi/2)=0,y^'(pi/2)=1 is y=t-pi/2