What is the solution for (2v+1)/(v-3v^2)dv= 1/x dx ?

The solution for (2v+1)/(v-3v^2)dv= 1/x dx is x=(3c_{1}v)/(3v(-3v+1)^{2/3)-(-3v+1)^{2/3}}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(2v+1)/(v-3v^2)dv= 1/x dx

\frac{2 v + 1}{v - 3 v ^{2}} d v = \frac{1}{x} d x

x = \frac{3 c _{1} v}{3 v ( - 3 v + 1 ) ^{\frac{2}{3}} - ( - 3 v + 1 ) ^{\frac{2}{3}}}

Solution steps

\frac{2 v + 1}{v - 3 v ^{2}} d v = \frac{1}{x} d x

Solve separable ODE:

x = \frac{3 c _{1} v}{3 v ( - 3 v + 1 ) ^{\frac{2}{3}} - ( - 3 v + 1 ) ^{\frac{2}{3}}}

x = \frac{3 c _{1} v}{3 v ( - 3 v + 1 ) ^{\frac{2}{3}} - ( - 3 v + 1 ) ^{\frac{2}{3}}}

y^{^{'}} sec (t) = - \frac{y ^{3}}{2}

\frac{d}{d x} (θ (x)) = sin (θ (x)) \cdot cos (θ (x))

3 x^{2} y^{5} d x + 5 x^{3} y^{4} d y = 0

(1 + cos (x)) d y = y \cdot sin (x) d x

(y - 2) \frac{d y}{d x} = 3 x - 5

What is the solution for (2v+1)/(v-3v^2)dv= 1/x dx ?
The solution for (2v+1)/(v-3v^2)dv= 1/x dx is x=(3c_{1}v)/(3v(-3v+1)^{2/3)-(-3v+1)^{2/3}}