What is the solution for y^{''}-2y^'-3y=2ex-10 ?

The solution for y^{''}-2y^'-3y=2ex-10 is y=c_{1}e^{3x}+c_{2}e^{-x}-(2ex)/3+(4e+30)/9

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y^{''}-2y^'-3y=2ex-10

y^{^{''}} - 2 y^{^{'}} - 3 y = 2 e x - 10

y = c_{1} e^{3 x} + c_{2} e^{- x} - \frac{2 e x}{3} + \frac{4 e + 30}{9}

Solution steps

y^{''} (x) - 2 y^{'} (x) - 3 y = 2 e x - 10

Solve linear ODE:

y = c_{1} e^{3 x} + c_{2} e^{- x} - \frac{2 e x}{3} + \frac{4 e + 30}{9}

y = c_{1} e^{3 x} + c_{2} e^{- x} - \frac{2 e x}{3} + \frac{4 e + 30}{9}

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

y^{^{''}} + 4 y = 4 sin (2 t) + 8 cos (2 t)

y^{^{''}} + 4 y = 2 cos (x)

y^{^{''}} + 7 y = 5 t^{2} - 5, y (0) = 0, y^{^{'}} (0) = - 2

y^{^{'''}} + y^{^{'}} = sec^{2} (x)

What is the solution for y^{''}-2y^'-3y=2ex-10 ?
The solution for y^{''}-2y^'-3y=2ex-10 is y=c_{1}e^{3x}+c_{2}e^{-x}-(2ex)/3+(4e+30)/9