What is the solution for 4y^{''}+4y^'+y=3xe^x ?

The solution for 4y^{''}+4y^'+y=3xe^x is y=c_{1}e^{-x/2}+c_{2}e^{-x/2}x+(e^xx)/3-4/9 e^x

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4y^{''}+4y^'+y=3xe^x

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

y = c_{1} e^{- \frac{x}{2}} + c_{2} e^{- \frac{x}{2}} x + \frac{e ^{x} x}{3} - \frac{4}{9} e^{x}

Solution steps

4 y^{''} (x) + 4 y^{'} (x) + y = 3 x e^{x}

Solve linear ODE:

y = c_{1} e^{- \frac{x}{2}} + c_{2} e^{- \frac{x}{2}} x + \frac{e ^{x} x}{3} - \frac{4}{9} e^{x}

y = c_{1} e^{- \frac{x}{2}} + c_{2} e^{- \frac{x}{2}} x + \frac{e ^{x} x}{3} - \frac{4}{9} e^{x}

y^{^{''}} - 6 y + 9 = e^{3 x}

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

y^{^{''}} + 81 y = 2 sin (9 t)

y^{^{''}} - 4 y = 12 x, y (0) = 4, y^{^{'}} (0) = 1

y^{^{''}} - 6 y^{^{'}} + 9 y = x

What is the solution for 4y^{''}+4y^'+y=3xe^x ?
The solution for 4y^{''}+4y^'+y=3xe^x is y=c_{1}e^{-x/2}+c_{2}e^{-x/2}x+(e^xx)/3-4/9 e^x