What is the solution for y^{''}+3y^'-10y=e^{-5x} ?

The solution for y^{''}+3y^'-10y=e^{-5x} is y=c_{1}e^{2x}+c_{2}e^{-5x}-(e^{-5x}x)/7

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y^{''}+3y^'-10y=e^{-5x}

y^{^{''}} + 3 y^{^{'}} - 10 y = e^{- 5 x}

y = c_{1} e^{2 x} + c_{2} e^{- 5 x} - \frac{e ^{- 5 x} x}{7}

Solution steps

y^{''} (x) + 3 y^{'} (x) - 10 y = e^{- 5 x}

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{- 5 x} - \frac{e ^{- 5 x} x}{7}

y = c_{1} e^{2 x} + c_{2} e^{- 5 x} - \frac{e ^{- 5 x} x}{7}

4 y^{^{''}} - 4 y^{^{'}} + y = e^{\frac{x}{2}} 1 - x

y^{^{''}} - y^{^{'}} = x^{2} e^{x}

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

y^{^{''}} - 4 y^{^{'}} + 5 y = e^{3 x} sin (2 x)

x^{^{''}} + x = tan (t)

What is the solution for y^{''}+3y^'-10y=e^{-5x} ?
The solution for y^{''}+3y^'-10y=e^{-5x} is y=c_{1}e^{2x}+c_{2}e^{-5x}-(e^{-5x}x)/7