What is the solution for 2y^{''}+3y^'-2y=25e^{-2x} ?

The solution for 2y^{''}+3y^'-2y=25e^{-2x} is y=c_{1}e^{x/2}+c_{2}e^{-2x}-5xe^{-2x}

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2y^{''}+3y^'-2y=25e^{-2x}

2 y^{^{''}} + 3 y^{^{'}} - 2 y = 25 e^{- 2 x}

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

Solution steps

2 y^{''} (x) + 3 y^{'} (x) - 2 y = 25 e^{- 2 x}

Solve linear ODE:

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

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

(- 2, - 1), (2, 4)

arctan (6 x^{2} - 1)

\int 8 x \cdot csc^{2} (4 x) d x

\int 2^{5 x + 2} d x

2 x y^{^{'}} + y = 6 x, y (4) = 15

What is the solution for 2y^{''}+3y^'-2y=25e^{-2x} ?
The solution for 2y^{''}+3y^'-2y=25e^{-2x} is y=c_{1}e^{x/2}+c_{2}e^{-2x}-5xe^{-2x}