What is the solution for 2y^{''}+y^'=3x^2+10x ?

The solution for 2y^{''}+y^'=3x^2+10x is y=c_{1}+c_{2}e^{-x/2}+x^3-x^2+4x

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2y^{''}+y^'=3x^2+10x

2 y^{^{''}} + y^{^{'}} = 3 x^{2} + 10 x

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

Solution steps

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

Solve linear ODE:

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

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

3 y^{^{''}} + 2 (9.81)^{2} y = 49 (9.81) cos (8 t)

y^{^{''}} + 2 y^{^{'}} + 5 y = 9 sin (2 t)

y^{^{''}} - 4 y^{^{'}} = 2 x - 1

y^{^{'''}} + 2 y^{^{''}} - 9 y^{^{'}} - 18 = 10 + 20 sin (5 t)

y^{^{'''}} - 3 y^{^{''}} + 4 y = x e^{2 x}

What is the solution for 2y^{''}+y^'=3x^2+10x ?
The solution for 2y^{''}+y^'=3x^2+10x is y=c_{1}+c_{2}e^{-x/2}+x^3-x^2+4x