What is the solution for y^'+8y-6x=0 ?

The solution for y^'+8y-6x=0 is y=(3(8x-1))/(32)+(c_{1})/(e^{8x)}

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y^'+8y-6x=0

y^{^{'}} + 8 y - 6 x = 0

y = \frac{3 ( 8 x - 1 )}{32} + \frac{c _{1}}{e ^{8 x}}

Solution steps

y^{'} (x) + 8 y - 6 x = 0

Solve linear ODE:

y = \frac{3 ( 8 x - 1 )}{32} + \frac{c _{1}}{e ^{8 x}}

y = \frac{3 ( 8 x - 1 )}{32} + \frac{c _{1}}{e ^{8 x}}

y^{^{'}} = \frac{y}{y - x}

n = 1 \sum \infty n^{- 2.4}

\int \frac{1}{1 + y ^{2}} d y

\int_{1}^{4} 8 x - 5 d x

\frac{\partial}{\partial x} (e^{x})

What is the solution for y^'+8y-6x=0 ?
The solution for y^'+8y-6x=0 is y=(3(8x-1))/(32)+(c_{1})/(e^{8x)}