What is the solution for y^{''}+4y+5=0 ?

The solution for y^{''}+4y+5=0 is y=c_{1}cos(2t)+c_{2}sin(2t)-5/4

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y^{''}+4y+5=0

y^{^{''}} + 4 y + 5 = 0

y = c_{1} cos (2 t) + c_{2} sin (2 t) - \frac{5}{4}

Solution steps

y^{''} (t) + 4 y + 5 = 0

Solve linear ODE:

y = c_{1} cos (2 t) + c_{2} sin (2 t) - \frac{5}{4}

y = c_{1} cos (2 t) + c_{2} sin (2 t) - \frac{5}{4}

- y^{^{''}} + y^{^{'}} - 6 y = 2 x

x^{^{''}} + 9 x = sin (t) + sin (3 t)

y^{^{'''}} - 3 y^{^{''}} + 3 y^{^{'}} - y = x - 5 e^{x}

y^{^{''}} - 10 y + 25 = 0

y^{^{''}} - 6 y^{^{'}} + 9 y = \frac{1}{x ^{2}} e^{3 x}

What is the solution for y^{''}+4y+5=0 ?
The solution for y^{''}+4y+5=0 is y=c_{1}cos(2t)+c_{2}sin(2t)-5/4