What is the solution for (dy)/(dx)y=u^2+2u,u= 1/(x^2) ?

The solution for (dy)/(dx)y=u^2+2u,u= 1/(x^2) is y=sqrt(2(u^2+2u)x+c_{1)},y=-sqrt(2(u^2+2u)x+c_{1)}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(dy)/(dx)y=u^2+2u,u= 1/(x^2)

\frac{d y}{d x} y = u^{2} + 2 u, u = \frac{1}{x ^{2}}

y = 2 (u^{2} + 2 u) x + c_{1}, y = - 2 (u^{2} + 2 u) x + c_{1}

Solution steps

\frac{d y}{d x} y = u^{2} + 2 u

Solve separable ODE:

y = 2 (u^{2} + 2 u) x + c_{1}, y = - 2 (u^{2} + 2 u) x + c_{1}

y = 2 (u^{2} + 2 u) x + c_{1}, y = - 2 (u^{2} + 2 u) x + c_{1}

d x + sin (y) x d y - sin (y) d y = 0

f^{^{'}} (x) = - 10 f (x)

\frac{d}{d x} ((f (x)) f (x)) = cos (2 x)

\frac{x d x}{( 1 + x )} = \frac{d y}{y ^{3}}

y^{^{'}} - 2 y y^{^{'}} + e^{x} = 0

What is the solution for (dy)/(dx)y=u^2+2u,u= 1/(x^2) ?
The solution for (dy)/(dx)y=u^2+2u,u= 1/(x^2) is y=sqrt(2(u^2+2u)x+c_{1)},y=-sqrt(2(u^2+2u)x+c_{1)}