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

The solution for 3y^{''}-2y^'=2x^2-x is y=c_{1}e^{(2x)/3}+c_{2}-(x^3)/3-(5x^2)/4-(15x)/4

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

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

y = c_{1} e^{\frac{2 x}{3}} + c_{2} - \frac{x ^{3}}{3} - \frac{5 x ^{2}}{4} - \frac{15 x}{4}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{\frac{2 x}{3}} + c_{2} - \frac{x ^{3}}{3} - \frac{5 x ^{2}}{4} - \frac{15 x}{4}

y = c_{1} e^{\frac{2 x}{3}} + c_{2} - \frac{x ^{3}}{3} - \frac{5 x ^{2}}{4} - \frac{15 x}{4}

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

y^{^{''}} - 17 y^{^{'}} + 72 y = 55 + 72 x

y^{^{''}} - y^{^{'}} - 6 y = 5 e^{2 x}

y^{^{''}} + 2 y^{^{'}} + 7 y = 6 cos (2 x)

y^{^{'''}} + 16 y^{^{'}} = sec (4 x)

What is the solution for 3y^{''}-2y^'=2x^2-x ?
The solution for 3y^{''}-2y^'=2x^2-x is y=c_{1}e^{(2x)/3}+c_{2}-(x^3)/3-(5x^2)/4-(15x)/4