What is the solution for y^{''}-4y^'+3y=(2x+1)e^{-x} ?

The solution for y^{''}-4y^'+3y=(2x+1)e^{-x} is y=c_{1}e^{3x}+c_{2}e^x+(e^{-x}x)/4+5/16 e^{-x}

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y^{''}-4y^'+3y=(2x+1)e^{-x}

y^{^{''}} - 4 y^{^{'}} + 3 y = (2 x + 1) e^{- x}

y = c_{1} e^{3 x} + c_{2} e^{x} + \frac{e ^{- x} x}{4} + \frac{5}{16} e^{- x}

Solution steps

y^{''} (x) - 4 y^{'} (x) + 3 y = (2 x + 1) e^{- x}

Solve linear ODE:

y = c_{1} e^{3 x} + c_{2} e^{x} + \frac{e ^{- x} x}{4} + \frac{5}{16} e^{- x}

y = c_{1} e^{3 x} + c_{2} e^{x} + \frac{e ^{- x} x}{4} + \frac{5}{16} e^{- x}

x \to 0 - lim (\frac{0}{x ^{2} ∣ x ∣})

\frac{d}{d x} (\frac{e ^{x}}{3})

\int_{4 y^{2} + 4 z^{2}}^{4} x d x

\frac{d}{d x} ((x^{2} - 1) (x^{3} + 3 x))

(y^{2})^{^{'}}

What is the solution for y^{''}-4y^'+3y=(2x+1)e^{-x} ?
The solution for y^{''}-4y^'+3y=(2x+1)e^{-x} is y=c_{1}e^{3x}+c_{2}e^x+(e^{-x}x)/4+5/16 e^{-x}