What is the solution for y^{''}+2y=-3 ?

The solution for y^{''}+2y=-3 is y=c_{1}cos(sqrt(2)t)+c_{2}sin(sqrt(2)t)-3/2

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y^{''}+2y=-3

y^{^{''}} + 2 y = - 3

y = c_{1} cos (2 t) + c_{2} sin (2 t) - \frac{3}{2}

Solution steps

y^{''} (t) + 2 y = - 3

Solve linear ODE:

y = c_{1} cos (2 t) + c_{2} sin (2 t) - \frac{3}{2}

y = c_{1} cos (2 t) + c_{2} sin (2 t) - \frac{3}{2}

y^{^{''}} + 8 y = 2 t^{2} - 2, y (0) = 0, y^{^{'}} (0) = - 5

y^{^{''}} + 25 y = 64 cos (3 x), y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} + y^{^{'}} - 2 y = - 2 x^{3} - 3 x^{2} + 8 x + 1

y^{^{''}} - 2 y^{^{'}} + 4 y = 4

y^{^{''}} + 2 y^{^{'}} = c

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