What is the solution for 2y^{''}+7y^'+5y=0 ?

The solution for 2y^{''}+7y^'+5y=0 is y=c_{1}e^{-t}+c_{2}e^{-(5t)/2}

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2y^{''}+7y^'+5y=0

2 y^{^{''}} + 7 y^{^{'}} + 5 y = 0

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

Solution steps

2 y^{''} (t) + 7 y^{'} (t) + 5 y = 0

Solve linear ODE:

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

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

\int_{3}^{5} (- x^{2} + 8 x - 12) d x

x \to 2 lim (3 x + 6)

\int (x^{2} + 1) (x^{3} + 3 x) d x

\frac{d}{d x} (8 e^{2})

n = 1 \sum \infty \frac{1}{n ^{\frac{3}{2}}}

What is the solution for 2y^{''}+7y^'+5y=0 ?
The solution for 2y^{''}+7y^'+5y=0 is y=c_{1}e^{-t}+c_{2}e^{-(5t)/2}