What is the solution for y^{''}-2y^'-15y=3 ?

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

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y^{''}-2y^'-15y=3

y^{^{''}} - 2 y^{^{'}} - 15 y = 3

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

Solution steps

y^{''} (t) - 2 y^{'} (t) - 15 y = 3

Solve linear ODE:

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

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

\frac{\partial}{\partial x} (x^{\frac{1}{2}} \cdot y)

\frac{d}{d x} ((x - \frac{2}{x})^{4})

f (x) = 2 x^{2}, at x = 0

\int 4 x^{- 3} d x

\int \frac{1}{( x + 1 ) ( x - 1 )} d x

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