What is the solution for y^'e^{-12x}+12e^{-12x}y=-6e^{-8x} ?

The solution for y^'e^{-12x}+12e^{-12x}y=-6e^{-8x} is y=-(3e^{4x})/8+(c_{1})/(e^{12x)}

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y^'e^{-12x}+12e^{-12x}y=-6e^{-8x}

y^{^{'}} e^{- 12 x} + 12 e^{- 12 x} y = - 6 e^{- 8 x}

y = - \frac{3 e ^{4 x}}{8} + \frac{c _{1}}{e ^{12 x}}

Solution steps

y^{'} (x) e^{- 12 x} + 12 e^{- 12 x} y = - 6 e^{- 8 x}

Solve linear ODE:

y = - \frac{3 e ^{4 x}}{8} + \frac{c _{1}}{e ^{12 x}}

y = - \frac{3 e ^{4 x}}{8} + \frac{c _{1}}{e ^{12 x}}

e^{- 2 t} sin (2) (t)

\frac{\partial}{\partial x} (\frac{3 x y ^{2}}{2})

\frac{d}{d x} (x^{2} - 2 x - 8)

x y^{^{'}} + y = 9 x y^{2}, y (1) = 2

x^{3} - 6 x, - 2 x

What is the solution for y^'e^{-12x}+12e^{-12x}y=-6e^{-8x} ?
The solution for y^'e^{-12x}+12e^{-12x}y=-6e^{-8x} is y=-(3e^{4x})/8+(c_{1})/(e^{12x)}