What is the solution for y^{''}+4y=-5,y(pi/8)=-1/4 ,y^'(pi/8)=2 ?

The solution for y^{''}+4y=-5,y(pi/8)=-1/4 ,y^'(pi/8)=2 is y=sqrt(2)sin(2t)-5/4

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y^{''}+4y=-5,y(pi/8)=-1/4 ,y^'(pi/8)=2

y^{^{''}} + 4 y = - 5, y (\frac{π}{8}) = - \frac{1}{4}, y^{^{'}} (\frac{π}{8}) = 2

y = 2 sin (2 t) - \frac{5}{4}

Solution steps

y^{''} (t) + 4 y = - 5

Solve linear ODE:

y = 2 sin (2 t) - \frac{5}{4}

y = 2 sin (2 t) - \frac{5}{4}

y^{^{''}} + 7 y^{^{'}} + 6 y = 3 sin (2 t)

y^{^{''}} - 2 y^{^{'}} + 5 y = 2 x^{2} + 12

y^{^{''}} + 2 y^{^{'}} + y = 5 e^{- t}

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

y^{^{''}} + 2 y^{^{'}} + 5 y = 50 x sin (x) + 100 x cos (x)

What is the solution for y^{''}+4y=-5,y(pi/8)=-1/4 ,y^'(pi/8)=2 ?
The solution for y^{''}+4y=-5,y(pi/8)=-1/4 ,y^'(pi/8)=2 is y=sqrt(2)sin(2t)-5/4