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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2txdx+(t^2-x^2)dt=0

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

x = - t^{2} + c_{1} t, x = - - t^{2} + c_{1} t

Solution steps

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

Solve Bernoulli ODE:

x = - t^{2} + c_{1} t, x = - - t^{2} + c_{1} t

x = - t^{2} + c_{1} t, x = - - t^{2} + c_{1} t

y = \frac{9}{x}

\int_{0}^{2} \frac{1}{2} d x

5 x - 4 x

x \to 3 - lim (x^{2} + 2 x)

\int e^{s t} d t

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