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

The solution for y^{''}-2y^'-15y=-2 is y=c_{1}e^{5t}+c_{2}e^{-3t}+2/15

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

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

y = c_{1} e^{5 t} + c_{2} e^{- 3 t} + \frac{2}{15}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{5 t} + c_{2} e^{- 3 t} + \frac{2}{15}

y = c_{1} e^{5 t} + c_{2} e^{- 3 t} + \frac{2}{15}

\frac{\partial}{\partial x} (\frac{3 y ^{2} - x ^{2}}{y ^{5}})

\frac{\partial}{\partial y} (e^{y} (x^{2} - y^{2}))

\frac{d}{d x} (\frac{4 x ^{2} + 8 x + 4}{x})

g (u) = 7 u + 5 u

tan (x^{π})

What is the solution for y^{''}-2y^'-15y=-2 ?
The solution for y^{''}-2y^'-15y=-2 is y=c_{1}e^{5t}+c_{2}e^{-3t}+2/15