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^{4x}+c_{2}e^{2x}+1/8 e^{-2x}+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^{4 x} + c_{2} e^{2 x} + \frac{1}{8} e^{- 2 x} + \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^{4 x} + c_{2} e^{2 x} + \frac{1}{8} e^{- 2 x} + \frac{x}{4} + \frac{3}{16}

y = c_{1} e^{4 x} + c_{2} e^{2 x} + \frac{1}{8} e^{- 2 x} + \frac{x}{4} + \frac{3}{16}

y^{^{''}} + 4 y = 8 cos (2 x) + 12 sin (2 x)

x^{^{''}} + 4 x^{^{'}} + 3 x = 1

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

y^{^{''}} - 2 y^{^{'}} - 3 y = 3 t e^{2 t}, y (0) = 3, y^{^{'}} (0) = 0

y^{^{''}} + 36 y = 10 cos (6 x) + 15 sin (6 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^{4x}+c_{2}e^{2x}+1/8 e^{-2x}+x/4+3/16