What is the solution for y^{''}+4y=-6,y(pi/8)=-1/2 ,y^'(pi/8)=2 ?

The solution for y^{''}+4y=-6,y(pi/8)=-1/2 ,y^'(pi/8)=2 is y=sqrt(2)sin(2t)-3/2

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y^{''}+4y=-6,y(pi/8)=-1/2 ,y^'(pi/8)=2

y^{^{''}} + 4 y = - 6, y (\frac{π}{8}) = - \frac{1}{2}, y^{^{'}} (\frac{π}{8}) = 2

y = 2 sin (2 t) - \frac{3}{2}

Solution steps

y^{''} (t) + 4 y = - 6

Solve linear ODE:

y = 2 sin (2 t) - \frac{3}{2}

y = 2 sin (2 t) - \frac{3}{2}

What is the solution for y^{''}+4y=-6,y(pi/8)=-1/2 ,y^'(pi/8)=2 ?
The solution for y^{''}+4y=-6,y(pi/8)=-1/2 ,y^'(pi/8)=2 is y=sqrt(2)sin(2t)-3/2