What is the solution for 2y^{''}-5y^'-12y=0 ?

The solution for 2y^{''}-5y^'-12y=0 is y=c_{1}e^{4t}+c_{2}e^{-(3t)/2}

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2y^{''}-5y^'-12y=0

2 y^{^{''}} - 5 y^{^{'}} - 12 y = 0

y = c_{1} e^{4 t} + c_{2} e^{- \frac{3 t}{2}}

Solution steps

2 y^{''} (t) - 5 y^{'} (t) - 12 y = 0

Solve linear ODE:

y = c_{1} e^{4 t} + c_{2} e^{- \frac{3 t}{2}}

y = c_{1} e^{4 t} + c_{2} e^{- \frac{3 t}{2}}

\int_{0}^{49} \frac{1}{49 - x} d x

x sin (\frac{5}{x})

\frac{d}{d x} (\frac{x}{1 - ln ( x - 9 )})

x y^{^{'}} - y = \frac{5}{4} x ln (x)

y = 8 x arccos (x - 1), at x = 1

What is the solution for 2y^{''}-5y^'-12y=0 ?
The solution for 2y^{''}-5y^'-12y=0 is y=c_{1}e^{4t}+c_{2}e^{-(3t)/2}