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}}

\int \frac{( x + 6 )}{x} d x

\frac{d}{d x} (x cos (x) + x^{2} sin (x))

\int 3 x + 1 d x

\int_{0}^{1} x e^{1 - x^{2}} d x

(3 x - 1 + 4 5 - 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)}