What is the solution for y^{''}-2y^'-3y=8ex-3 ?

The solution for y^{''}-2y^'-3y=8ex-3 is y=c_{1}e^{3x}+c_{2}e^{-x}-(8ex)/3+(16e+9)/9

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y^{''}-2y^'-3y=8ex-3

y^{^{''}} - 2 y^{^{'}} - 3 y = 8 e x - 3

y = c_{1} e^{3 x} + c_{2} e^{- x} - \frac{8 e x}{3} + \frac{16 e + 9}{9}

Solution steps

y^{''} (x) - 2 y^{'} (x) - 3 y = 8 e x - 3

Solve linear ODE:

y = c_{1} e^{3 x} + c_{2} e^{- x} - \frac{8 e x}{3} + \frac{16 e + 9}{9}

y = c_{1} e^{3 x} + c_{2} e^{- x} - \frac{8 e x}{3} + \frac{16 e + 9}{9}

y^{^{''}} + 2 y^{^{'}} + 4 y = 6 cos (3 x)

x^{^{''}} + 4 x^{^{'}} + 3 x = 1, x^{^{'}} (0) = 0

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

\frac{d ^{2} y}{d x ^{2}} + y = 2 e^{3 x}

y^{^{'''}} - y^{^{''}} - 4 y^{^{'}} + 4 y = 10 e^{3 t}

What is the solution for y^{''}-2y^'-3y=8ex-3 ?
The solution for y^{''}-2y^'-3y=8ex-3 is y=c_{1}e^{3x}+c_{2}e^{-x}-(8ex)/3+(16e+9)/9