What is the solution for 2y^{''}+y^'-10y=0 ?

The solution for 2y^{''}+y^'-10y=0 is y=c_{1}e^{2t}+c_{2}e^{-(5t)/2}

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2y^{''}+y^'-10y=0

2 y^{^{''}} + y^{^{'}} - 10 y = 0

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

Solution steps

2 y^{''} (t) + y^{'} (t) - 10 y = 0

Solve linear ODE:

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

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

\int 3 e^{- 0.8 x} d x

\int 12 x - x^{2} d x

y = 4 x, y = 9 x^{2}

\frac{d}{d x} (ln (sech (2 x)))

What is the solution for 2y^{''}+y^'-10y=0 ?
The solution for 2y^{''}+y^'-10y=0 is y=c_{1}e^{2t}+c_{2}e^{-(5t)/2}