What is the solution for y^{''}-y^'=3-4x,y(0)=-3,y^'(0)=0 ?

The solution for y^{''}-y^'=3-4x,y(0)=-3,y^'(0)=0 is y=-e^x-2+2x^2+x

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y^{''}-y^'=3-4x,y(0)=-3,y^'(0)=0

y^{^{''}} - y^{^{'}} = 3 - 4 x, y (0) = - 3, y^{^{'}} (0) = 0

y = - e^{x} - 2 + 2 x^{2} + x

Solution steps

y^{''} (x) - y^{'} (x) = 3 - 4 x

Solve linear ODE:

y = - e^{x} - 2 + 2 x^{2} + x

y = - e^{x} - 2 + 2 x^{2} + x

y^{^{''}} - 4 y^{^{'}} + 4 y = e^{2 x} \cdot ln (x)

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

y^{^{''}} + 25 y = x^{2}

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

5 y^{^{''}} + 6 y^{^{'}} + \frac{5}{1.96} y = 3 cos (2 t)

What is the solution for y^{''}-y^'=3-4x,y(0)=-3,y^'(0)=0 ?
The solution for y^{''}-y^'=3-4x,y(0)=-3,y^'(0)=0 is y=-e^x-2+2x^2+x