What is the solution for y^{''}+7/2 y^'+3/2 y=2e^{-3x} ?

The solution for y^{''}+7/2 y^'+3/2 y=2e^{-3x} is y=c_{1}e^{-x/2}+c_{2}e^{-3x}-(4e^{-3x}x)/5

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y^{''}+7/2 y^'+3/2 y=2e^{-3x}

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

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

Solution steps

y^{''} (x) + \frac{7}{2} y^{'} (x) + \frac{3}{2} y = 2 e^{- 3 x}

Solve linear ODE:

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

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

y^{^{''}} + 4^{2} y = cos (2 t)

y^{^{''}} - 10 y^{^{'}} + 21 y = 192 e^{t}

y^{^{''}} - 6 y + 10 y = 10 x^{2}

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

y^{^{''}} + 14 y^{^{'}} + 49 y = x^{- 3} e^{- 7 x}

What is the solution for y^{''}+7/2 y^'+3/2 y=2e^{-3x} ?
The solution for y^{''}+7/2 y^'+3/2 y=2e^{-3x} is y=c_{1}e^{-x/2}+c_{2}e^{-3x}-(4e^{-3x}x)/5