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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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

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

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

so l v e f or d, d (t) = 4000 - 3000 e^{- 0.06 t}

(d^{2} - d - 2) \cdot y = 2 \cdot lo g_{10} (x) + \frac{1}{x} + \frac{1}{( x ^{2} )}

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