What is the solution for y^{''}+y^'=3x^2,y(0)=0,y^'(0)=1 ?

The solution for y^{''}+y^'=3x^2,y(0)=0,y^'(0)=1 is y=-5+5e^{-x}+x^3-3x^2+6x

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y^{''}+y^'=3x^2,y(0)=0,y^'(0)=1

y^{^{''}} + y^{^{'}} = 3 x^{2}, y (0) = 0, y^{^{'}} (0) = 1

y = - 5 + 5 e^{- x} + x^{3} - 3 x^{2} + 6 x

Solution steps

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

Solve linear ODE:

y = - 5 + 5 e^{- x} + x^{3} - 3 x^{2} + 6 x

y = - 5 + 5 e^{- x} + x^{3} - 3 x^{2} + 6 x

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

(D^{2} + 4) y = 8 cos^{2} (x)

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

y^{^{''}} + 4 y^{^{'}} + 4 y = e^{- x} cos (x)

y^{^{''}} + 3 y^{^{'}} - 4 y = e^{x}

What is the solution for y^{''}+y^'=3x^2,y(0)=0,y^'(0)=1 ?
The solution for y^{''}+y^'=3x^2,y(0)=0,y^'(0)=1 is y=-5+5e^{-x}+x^3-3x^2+6x