What is the solution for y^{''}+4y=-1,y(pi/8)= 3/4 ,y^'(pi/8)=2 ?

The solution for y^{''}+4y=-1,y(pi/8)= 3/4 ,y^'(pi/8)=2 is y=sqrt(2)sin(2t)-1/4

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y^{''}+4y=-1,y(pi/8)= 3/4 ,y^'(pi/8)=2

y^{^{''}} + 4 y = - 1, y (\frac{π}{8}) = \frac{3}{4}, y^{^{'}} (\frac{π}{8}) = 2

y = 2 sin (2 t) - \frac{1}{4}

Solution steps

y^{''} (t) + 4 y = - 1

Solve linear ODE:

y = 2 sin (2 t) - \frac{1}{4}

y = 2 sin (2 t) - \frac{1}{4}

y^{^{''}} + 5 y^{^{'}} + 4 y = 12

y^{^{''}} - 22 y^{^{'}} + 121 y = \frac{e ^{- 11 x}}{1 + x ^{2}}

x^{^{''}} - x = sin (t)

y^{^{''}} + 5 y^{^{'}} - 6 y = 10 e^{2 x}

y^{^{''}} + 4 y = 9 sin (t)

What is the solution for y^{''}+4y=-1,y(pi/8)= 3/4 ,y^'(pi/8)=2 ?
The solution for y^{''}+4y=-1,y(pi/8)= 3/4 ,y^'(pi/8)=2 is y=sqrt(2)sin(2t)-1/4