What is the solution for y^{'''}-5y^{''}=2x ?

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

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

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

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

Solution steps

y^{'''} (x) - 5 y^{''} (x) = 2 x

Solve linear ODE:

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

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

y^{^{''}} - 2 y^{^{'}} - 3 y = e t, y (0) = 2, y^{^{'}} (0) = 4

x^{^{''}} + 23.04 x = 2 cos (5 t)

y^{^{''}} + 4 y = 9, t y (0) = 0, y^{^{'}} (0) = 7

y^{^{''}} + 8 y^{^{'}} + 12 y = 1 - H (t)

y^{^{'''}} - 11 y^{^{''}} + 28 y^{^{'}} = 36 e^{x}, y (0) = 25

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