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

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

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8y^'=3e^{x/8}+y

8 y^{^{'}} = 3 e^{\frac{x}{8}} + y

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

Solution steps

8 y^{'} (x) = 3 e^{\frac{x}{8}} + y

Solve linear ODE:

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

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

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

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