What is the solution for y^{'''}+y^{''}=7x^2 ?

The solution for y^{'''}+y^{''}=7x^2 is y=c_{1}e^{-x}+c_{2}+c_{3}x+(7x^4}{12}-\frac{7x^3)/3+7x^2

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y^{'''}+y^{''}=7x^2

y^{^{'''}} + y^{^{''}} = 7 x^{2}

y = c_{1} e^{- x} + c_{2} + c_{3} x + \frac{7 x ^{4}}{12} - \frac{7 x ^{3}}{3} + 7 x^{2}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{- x} + c_{2} + c_{3} x + \frac{7 x ^{4}}{12} - \frac{7 x ^{3}}{3} + 7 x^{2}

y = c_{1} e^{- x} + c_{2} + c_{3} x + \frac{7 x ^{4}}{12} - \frac{7 x ^{3}}{3} + 7 x^{2}

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

y^{^{''}} + 9 y = sin (x) - 2

y^{^{'''}} - y^{^{'}} = tan (x)

y^{^{''}} + 4 y^{^{'}} + 8 y = 12 e^{- 2 x} sec (2 x)

y^{^{''}} - 4 y = 20

What is the solution for y^{'''}+y^{''}=7x^2 ?
The solution for y^{'''}+y^{''}=7x^2 is y=c_{1}e^{-x}+c_{2}+c_{3}x+(7x^4}{12}-\frac{7x^3)/3+7x^2