What is the solution for 2y^{''}-15y^'-8y=0 ?

The solution for 2y^{''}-15y^'-8y=0 is y=c_{1}e^{8t}+c_{2}e^{-t/2}

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2y^{''}-15y^'-8y=0

2 y^{^{''}} - 15 y^{^{'}} - 8 y = 0

y = c_{1} e^{8 t} + c_{2} e^{- \frac{t}{2}}

Solution steps

2 y^{''} (t) - 15 y^{'} (t) - 8 y = 0

Solve linear ODE:

y = c_{1} e^{8 t} + c_{2} e^{- \frac{t}{2}}

y = c_{1} e^{8 t} + c_{2} e^{- \frac{t}{2}}

\int_{- 1}^{1} x - 1 - x^{2} d x

y = (x^{6} - 10) x

\int sin^{5} (x) cos^{8} (x) d x

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

\frac{s}{( s ^{2} + 9 ) ^{2}}

What is the solution for 2y^{''}-15y^'-8y=0 ?
The solution for 2y^{''}-15y^'-8y=0 is y=c_{1}e^{8t}+c_{2}e^{-t/2}