What is the solution for y^{''}+3y^'=8x+5 ?

The solution for y^{''}+3y^'=8x+5 is y=c_{1}+c_{2}e^{-3x}+(4x^2)/3+(7x)/9

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y^{''}+3y^'=8x+5

y^{^{''}} + 3 y^{^{'}} = 8 x + 5

y = c_{1} + c_{2} e^{- 3 x} + \frac{4 x ^{2}}{3} + \frac{7 x}{9}

Solution steps

y^{''} (x) + 3 y^{'} (x) = 8 x + 5

Solve linear ODE:

y = c_{1} + c_{2} e^{- 3 x} + \frac{4 x ^{2}}{3} + \frac{7 x}{9}

y = c_{1} + c_{2} e^{- 3 x} + \frac{4 x ^{2}}{3} + \frac{7 x}{9}

y^{^{''}} + 2 y^{^{'}} + y = 4 e^{- x} ln (x)

3 y^{^{''}} + 8 y^{^{'}} - 3 y = 8 - 3 t

2 y^{^{''}} + 4 y = 6 t, y (0) = 4.9, y^{^{'}} (0) = 0

y^{^{''}} - 5 y^{^{'}} + 6 y = 24

D (t) = - 200 t + 9000

What is the solution for y^{''}+3y^'=8x+5 ?
The solution for y^{''}+3y^'=8x+5 is y=c_{1}+c_{2}e^{-3x}+(4x^2)/3+(7x)/9