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

The solution for y^{''}-2y^'-15y=3x is y=c_{1}e^{5x}+c_{2}e^{-3x}-x/5+2/75

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

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

y = c_{1} e^{5 x} + c_{2} e^{- 3 x} - \frac{x}{5} + \frac{2}{75}

Solution steps

y^{''} (x) - 2 y^{'} (x) - 15 y = 3 x

Solve linear ODE:

y = c_{1} e^{5 x} + c_{2} e^{- 3 x} - \frac{x}{5} + \frac{2}{75}

y = c_{1} e^{5 x} + c_{2} e^{- 3 x} - \frac{x}{5} + \frac{2}{75}

(y^{^{''}} + 2 y^{^{'}} + 5) = e^{\frac{x}{3}}

y^{^{''}} + 5 y = - 180 x^{2} e^{(5 x)}

y^{^{''}} - 10 y^{^{'}} + 21 y = e^{7 t}

y^{^{''}} + 7 y^{^{'}} + 10 = x

y^{^{''}} - 9 y = (x^{2} - 2) \cdot sin (3 x)

What is the solution for y^{''}-2y^'-15y=3x ?
The solution for y^{''}-2y^'-15y=3x is y=c_{1}e^{5x}+c_{2}e^{-3x}-x/5+2/75