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 e^{- \frac{x}{b}} d x

\int \frac{( x - 1 ) ^{2}}{x} d x

\frac{d}{d x} (20 x (1 - x)^{3})

\int \frac{1}{( 2 x + 5 ) ^{3}} d x

\int sin (2 x) x d x

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