What is the solution for y^{''}-3y^'=4x ?

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

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y^{''}-3y^'=4x

y^{^{''}} - 3 y^{^{'}} = 4 x

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

Solution steps

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

Solve linear ODE:

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

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

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

y^{^{''}} + y = e^{x} (6 - x)

x^{^{''}} + 6 x^{^{'}} + 9 x = 10 cos (2 t)

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

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

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