What is the solution for 6y^{''}-15y^'=3x^2 ?

The solution for 6y^{''}-15y^'=3x^2 is y=c_{1}e^{(5x)/2}+c_{2}-(x^3)/(15)-(2x^2)/(25)-(8x)/(125)

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6y^{''}-15y^'=3x^2

6 y^{^{''}} - 15 y^{^{'}} = 3 x^{2}

y = c_{1} e^{\frac{5 x}{2}} + c_{2} - \frac{x ^{3}}{15} - \frac{2 x ^{2}}{25} - \frac{8 x}{125}

Solution steps

6 y^{''} (x) - 15 y^{'} (x) = 3 x^{2}

Solve linear ODE:

y = c_{1} e^{\frac{5 x}{2}} + c_{2} - \frac{x ^{3}}{15} - \frac{2 x ^{2}}{25} - \frac{8 x}{125}

y = c_{1} e^{\frac{5 x}{2}} + c_{2} - \frac{x ^{3}}{15} - \frac{2 x ^{2}}{25} - \frac{8 x}{125}

y^{^{''}} + 14 y^{^{'}} + 49 y = t - 2 e - 7 t

y^{^{''}} + 9 y = (x^{2} + 1) e^{3 x}

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

3 x^{^{''}} + 12 x = sin (2 t)

y^{^{''}} + 25 y = 2 sin (2 x), y (0) = 1, y^{^{'}} (0) = 0

What is the solution for 6y^{''}-15y^'=3x^2 ?
The solution for 6y^{''}-15y^'=3x^2 is y=c_{1}e^{(5x)/2}+c_{2}-(x^3)/(15)-(2x^2)/(25)-(8x)/(125)