What is the solution for y^{''}+4y^1-y=0 ?

The solution for y^{''}+4y^1-y=0 is y=c_{1}cos(sqrt(3)t)+c_{2}sin(sqrt(3)t)

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y^{''}+4y^1-y=0

y^{^{''}} + 4 y^{1} - y = 0

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

Solution steps

y^{''} (t) + 4 y^{1} - y = 0

Solve linear ODE:

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

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

y^{^{''}} + π^{2} \cdot y = 0

y^{^{''}} + 16 y^{^{'}} + 100 y = 0, y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} + (\frac{25}{6}) y^{^{'}} + 25 y = 0

\frac{d ^{2} y}{d t ^{2}} + 16 y = 0, y (0) = 10, y^{^{'}} (0) = 0

(D^{3} - 1) y = 0

What is the solution for y^{''}+4y^1-y=0 ?
The solution for y^{''}+4y^1-y=0 is y=c_{1}cos(sqrt(3)t)+c_{2}sin(sqrt(3)t)