What is the solution for y^{''}+3y^'-10y=20+5e^{3x} ?

The solution for y^{''}+3y^'-10y=20+5e^{3x} is y=c_{1}e^{2x}+c_{2}e^{-5x}-2+5/8 e^{3x}

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y^{''}+3y^'-10y=20+5e^{3x}

y^{^{''}} + 3 y^{^{'}} - 10 y = 20 + 5 e^{3 x}

y = c_{1} e^{2 x} + c_{2} e^{- 5 x} - 2 + \frac{5}{8} e^{3 x}

Solution steps

y^{''} (x) + 3 y^{'} (x) - 10 y = 20 + 5 e^{3 x}

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{- 5 x} - 2 + \frac{5}{8} e^{3 x}

y = c_{1} e^{2 x} + c_{2} e^{- 5 x} - 2 + \frac{5}{8} e^{3 x}

x^{^{''}} + 4 x^{^{'}} + 14 x = 9 cos (3 t)

y^{^{''}} + 6 y^{^{'}} + 9 y = 4 sin (6 t)

y^{^{''}} + 9 y = t + 3, y (0) = 1, y^{^{'}} (0) = 2

y^{^{''}} - 5 y^{^{'}} + 2 y = 6 t^{2} + 2 t + 7

y^{^{''}} + 5 y^{^{'}} + 80 y = 220 cos (11 t)

What is the solution for y^{''}+3y^'-10y=20+5e^{3x} ?
The solution for y^{''}+3y^'-10y=20+5e^{3x} is y=c_{1}e^{2x}+c_{2}e^{-5x}-2+5/8 e^{3x}