What is the solution for y^{'''}+3y^{''}=-36x^2+12x+42 ?

The solution for y^{'''}+3y^{''}=-36x^2+12x+42 is y=c_{1}e^{-3x}+c_{2}+c_{3}x-x^4+2x^3+5x^2

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y^{'''}+3y^{''}=-36x^2+12x+42

y^{^{'''}} + 3 y^{^{''}} = - 36 x^{2} + 12 x + 42

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

Solution steps

y^{'''} (x) + 3 y^{''} (x) = - 36 x^{2} + 12 x + 42

Solve linear ODE:

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

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

y^{^{''}} - y^{^{'}} = e^{2 x} \cdot cos (e^{x})

y^{^{''}} + 5 y^{^{'}} + 4 y = 8 x^{2} + 3 + 2 cos^{2} (x)

y^{^{''}} - 3 y^{^{'}} + 10 y = cosh (2 x)

(d^{2} + 4) y = tan^{2} (x)

(d^{3} - 1) \cdot y = e^{x}

What is the solution for y^{'''}+3y^{''}=-36x^2+12x+42 ?
The solution for y^{'''}+3y^{''}=-36x^2+12x+42 is y=c_{1}e^{-3x}+c_{2}+c_{3}x-x^4+2x^3+5x^2