What is the solution for y^{''}+4y^'+3y=7e^{-x} ?

The solution for y^{''}+4y^'+3y=7e^{-x} is y=c_{1}e^{-x}+c_{2}e^{-3x}+(7e^{-x}x)/2

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y^{''}+4y^'+3y=7e^{-x}

y^{^{''}} + 4 y^{^{'}} + 3 y = 7 e^{- x}

y = c_{1} e^{- x} + c_{2} e^{- 3 x} + \frac{7 e ^{- x} x}{2}

Solution steps

y^{''} (x) + 4 y^{'} (x) + 3 y = 7 e^{- x}

Solve linear ODE:

y = c_{1} e^{- x} + c_{2} e^{- 3 x} + \frac{7 e ^{- x} x}{2}

y = c_{1} e^{- x} + c_{2} e^{- 3 x} + \frac{7 e ^{- x} x}{2}

y^{^{''}} + 8 y^{^{'}} + 16 y = e^{- 4 t}

y^{^{''}} + 9 y = sin (3 x) + e^{- x}

y^{^{''}} - y^{^{'}} - 2 y = cosh (2 x)

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

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

What is the solution for y^{''}+4y^'+3y=7e^{-x} ?
The solution for y^{''}+4y^'+3y=7e^{-x} is y=c_{1}e^{-x}+c_{2}e^{-3x}+(7e^{-x}x)/2