What is the solution for 8y^{''}+3y^'=12x ?

The solution for 8y^{''}+3y^'=12x is y=c_{1}+c_{2}e^{-(3x)/8}+2x^2-(32x)/3

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8y^{''}+3y^'=12x

8 y^{^{''}} + 3 y^{^{'}} = 12 x

y = c_{1} + c_{2} e^{- \frac{3 x}{8}} + 2 x^{2} - \frac{32 x}{3}

Solution steps

8 y^{''} (x) + 3 y^{'} (x) = 12 x

Solve linear ODE:

y = c_{1} + c_{2} e^{- \frac{3 x}{8}} + 2 x^{2} - \frac{32 x}{3}

y = c_{1} + c_{2} e^{- \frac{3 x}{8}} + 2 x^{2} - \frac{32 x}{3}

y^{^{''}} + y^{^{'}} - 6 y = e^{2 t}

y^{^{''}} - 4 y^{^{'}} + 5 y = (x + 1)^{3}

y^{^{''}} + 9 y = 18 t, y (\frac{π}{2}) = 0

y^{^{''}} + 16 y = csc^{2} (4 t)

y^{^{''}} + 6 y^{^{'}} + 9 y = e^{- 3 x}

What is the solution for 8y^{''}+3y^'=12x ?
The solution for 8y^{''}+3y^'=12x is y=c_{1}+c_{2}e^{-(3x)/8}+2x^2-(32x)/3