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

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

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

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

y = 2 sin (2 t) - \frac{3}{4}

Solution steps

y^{''} (t) + 4 y = - 3

Solve linear ODE:

y = 2 sin (2 t) - \frac{3}{4}

y = 2 sin (2 t) - \frac{3}{4}

y^{^{''}} + y^{^{'}} = e^{2 t} + 3 t e^{t}

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

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

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

\frac{d ^{2} x}{d t ^{2}} + x k + a = 0

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