What is the solution for 25y^{''}+60y^'+36y=0 ?

The solution for 25y^{''}+60y^'+36y=0 is y=c_{1}e^{-(6t)/5}+c_{2}e^{-(6t)/5}t

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25y^{''}+60y^'+36y=0

25 y^{^{''}} + 60 y^{^{'}} + 36 y = 0

y = c_{1} e^{- \frac{6 t}{5}} + c_{2} e^{- \frac{6 t}{5}} t

Solution steps

25 y^{''} (t) + 60 y^{'} (t) + 36 y = 0

Solve linear ODE:

y = c_{1} e^{- \frac{6 t}{5}} + c_{2} e^{- \frac{6 t}{5}} t

y = c_{1} e^{- \frac{6 t}{5}} + c_{2} e^{- \frac{6 t}{5}} t

y = \frac{x ^{2} + 7 x - 2}{x - 2}

\frac{d x}{d t} = 4 (x^{2} + 1), x (\frac{π}{4}) = 1

\int tan^{4} (2 x) d x

t^{5} y^{4} - t^{3} y^{^{'}} + 6 y = 0

y^{^{'}} = \frac{3 ( e ^{x} + 1 )}{2 y}

What is the solution for 25y^{''}+60y^'+36y=0 ?
The solution for 25y^{''}+60y^'+36y=0 is y=c_{1}e^{-(6t)/5}+c_{2}e^{-(6t)/5}t