What is the solution for y^{''}+2y^'-24y=0 ?

The solution for y^{''}+2y^'-24y=0 is y=c_{1}e^{4t}+c_{2}e^{-6t}

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y^{''}+2y^'-24y=0

y^{^{''}} + 2 y^{^{'}} - 24 y = 0

y = c_{1} e^{4 t} + c_{2} e^{- 6 t}

Solution steps

y^{''} (t) + 2 y^{'} (t) - 24 y = 0

Solve linear ODE:

y = c_{1} e^{4 t} + c_{2} e^{- 6 t}

y = c_{1} e^{4 t} + c_{2} e^{- 6 t}

x y + y^{^{'}} = 76 x

\int sin (9 x) d x

\int sin (4 θ) d θ

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

What is the solution for y^{''}+2y^'-24y=0 ?
The solution for y^{''}+2y^'-24y=0 is y=c_{1}e^{4t}+c_{2}e^{-6t}