What is the solution for (12-20y+e^{-5x})dx-4dy=0 ?

The solution for (12-20y+e^{-5x})dx-4dy=0 is y= x/(4e^{5x)}+3/5+(c_{1})/(e^{5x)}

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(12-20y+e^{-5x})dx-4dy=0

(12 - 20 y + e^{- 5 x}) d x - 4 d y = 0

y = \frac{x}{4 e ^{5 x}} + \frac{3}{5} + \frac{c _{1}}{e ^{5 x}}

Solution steps

(12 - 20 y + e^{- 5 x}) d x - 4 d y = 0

Solve linear ODE:

y = \frac{x}{4 e ^{5 x}} + \frac{3}{5} + \frac{c _{1}}{e ^{5 x}}

y = \frac{x}{4 e ^{5 x}} + \frac{3}{5} + \frac{c _{1}}{e ^{5 x}}

\int 8 x^{\frac{1}{3}} + 8 x^{- \frac{1}{3}} d x

\int_{0}^{2} ln (2 x) d x

\frac{\partial}{\partial x} (\frac{y}{ln ( x - 1 )})

\frac{d y}{d t} + a y = 0

2 y^{^{'}} + x y^{^{''}} = 0

What is the solution for (12-20y+e^{-5x})dx-4dy=0 ?
The solution for (12-20y+e^{-5x})dx-4dy=0 is y= x/(4e^{5x)}+3/5+(c_{1})/(e^{5x)}