What is the solution for y^{''}+4y=-2,y(pi/8)= 1/2 ,y^'(pi/8)=2 ?

The solution for y^{''}+4y=-2,y(pi/8)= 1/2 ,y^'(pi/8)=2 is y=sqrt(2)sin(2t)-1/2

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y^{''}+4y=-2,y(pi/8)= 1/2 ,y^'(pi/8)=2

y^{^{''}} + 4 y = - 2, y (\frac{π}{8}) = \frac{1}{2}, y^{^{'}} (\frac{π}{8}) = 2

y = 2 sin (2 t) - \frac{1}{2}

Solution steps

y^{''} (t) + 4 y = - 2

Solve linear ODE:

y = 2 sin (2 t) - \frac{1}{2}

y = 2 sin (2 t) - \frac{1}{2}

y^{^{''}} - 4 y^{^{'}} - 5 y = e^{x} + 2 x

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

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

y^{^{'''}} + 2 y^{^{''}} = 24 x^{2} + 24 x + 4

y^{^{''}} + 5 y^{^{'}} + 6 y = 8 e^{5 t}

What is the solution for y^{''}+4y=-2,y(pi/8)= 1/2 ,y^'(pi/8)=2 ?
The solution for y^{''}+4y=-2,y(pi/8)= 1/2 ,y^'(pi/8)=2 is y=sqrt(2)sin(2t)-1/2