What is the solution for (12-6y+e^{-3x})dx-2dy=0 ?

The solution for (12-6y+e^{-3x})dx-2dy=0 is y= x/(2e^{3x)}+2+(c_{1})/(e^{3x)}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(12-6y+e^{-3x})dx-2dy=0

(12 - 6 y + e^{- 3 x}) d x - 2 d y = 0

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

Solution steps

(12 - 6 y + e^{- 3 x}) d x - 2 d y = 0

Solve linear ODE:

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

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

\int \frac{x ^{3}}{x ^{4} + 10} d x

(sin^{3} (t))^{^{'}}

y^{^{''}} = - 4 y + e^{x}

\int \frac{x ^{2} - 3 x - 8}{x ^{2} - 2 x + 1} d x

x y - 6 y^{2} = 4, (14, 2)

What is the solution for (12-6y+e^{-3x})dx-2dy=0 ?
The solution for (12-6y+e^{-3x})dx-2dy=0 is y= x/(2e^{3x)}+2+(c_{1})/(e^{3x)}