What is the solution for (10-6y+e^{-3x})dx-2dy=0 ?

The solution for (10-6y+e^{-3x})dx-2dy=0 is y= x/(2e^{3x)}+5/3+(c_{1})/(e^{3x)}

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(10-6y+e^{-3x})dx-2dy=0

(10 - 6 y + e^{- 3 x}) d x - 2 d y = 0

y = \frac{x}{2 e ^{3 x}} + \frac{5}{3} + \frac{c _{1}}{e ^{3 x}}

Solution steps

(10 - 6 y + e^{- 3 x}) d x - 2 d y = 0

Solve linear ODE:

y = \frac{x}{2 e ^{3 x}} + \frac{5}{3} + \frac{c _{1}}{e ^{3 x}}

y = \frac{x}{2 e ^{3 x}} + \frac{5}{3} + \frac{c _{1}}{e ^{3 x}}

\frac{d}{d x} (\frac{x ^{2} + 1}{x ^{2} + x + 1})

\int arcsin (\frac{x}{x + 1}) d x

\frac{\partial}{\partial x} (e^{- \frac{1}{2} x y})

\int - 2 x + 6 d x

y^{^{'}} + \frac{1}{x + 7} y = x^{- 2}, y (1) = 4

What is the solution for (10-6y+e^{-3x})dx-2dy=0 ?
The solution for (10-6y+e^{-3x})dx-2dy=0 is y= x/(2e^{3x)}+5/3+(c_{1})/(e^{3x)}