What is the solution for (2x+t)(dx)/(dt)+(x-2t)=0 ?

The solution for (2x+t)(dx)/(dt)+(x-2t)=0 is x=(-t+sqrt(5t^2+c_{1)})/2 ,x=(-t-sqrt(5t^2+c_{1)})/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(2x+t)(dx)/(dt)+(x-2t)=0

(2 x + t) \frac{d x}{d t} + (x - 2 t) = 0

x = \frac{- t + 5 t ^{2} + c _{1}}{2}, x = \frac{- t - 5 t ^{2} + c _{1}}{2}

Solution steps

(2 x + t) \frac{d x}{d t} + (x - 2 t) = 0

Solve exact ODE:

x = \frac{- t + 5 t ^{2} + c _{1}}{2}, x = \frac{- t - 5 t ^{2} + c _{1}}{2}

x = \frac{- t + 5 t ^{2} + c _{1}}{2}, x = \frac{- t - 5 t ^{2} + c _{1}}{2}

\int - 2 csc^{2} (x) d x

x \to 0 lim (x^{3} \cdot csc^{2} (x))

f (x) = sin^{5} (2 x) cos^{3} (2 x)

(x^{2} - 1) y^{^{'}} = 2 y

y = \frac{x ^{2} + 1}{3 x + 1}, at x = 0

What is the solution for (2x+t)(dx)/(dt)+(x-2t)=0 ?
The solution for (2x+t)(dx)/(dt)+(x-2t)=0 is x=(-t+sqrt(5t^2+c_{1)})/2 ,x=(-t-sqrt(5t^2+c_{1)})/2