What is the solution for y^{''}+6y^'+8y=3e^{-2x}+2x ?

The solution for y^{''}+6y^'+8y=3e^{-2x}+2x is y=c_{1}e^{-2x}+c_{2}e^{-4x}+(3e^{-2x}x)/2+x/4-3/16

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y^{''}+6y^'+8y=3e^{-2x}+2x

y^{^{''}} + 6 y^{^{'}} + 8 y = 3 e^{- 2 x} + 2 x

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

Solution steps

y^{''} (x) + 6 y^{'} (x) + 8 y = 3 e^{- 2 x} + 2 x

Solve linear ODE:

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

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

\frac{\partial}{\partial x} (2 x)

\int 2 cos (2 θ) d θ

(2, 59), (4, 91)

f (t) = sin (3 t)

\frac{\partial}{\partial x} (e^{x} cos (y) + x)

What is the solution for y^{''}+6y^'+8y=3e^{-2x}+2x ?
The solution for y^{''}+6y^'+8y=3e^{-2x}+2x is y=c_{1}e^{-2x}+c_{2}e^{-4x}+(3e^{-2x}x)/2+x/4-3/16