What is the solution for y^'= x/y+(2y)/x ?

The solution for y^'= x/y+(2y)/x is y=xsqrt(-1+c_{1)x^2},y=-xsqrt(-1+c_{1)x^2}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

y^'= x/y+(2y)/x

y^{^{'}} = \frac{x}{y} + \frac{2 y}{x}

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

Solution steps

y^{'} (x) = \frac{x}{y} + \frac{2 y}{x}

Solve Bernoulli ODE:

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

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

\int \frac{1}{4 x ^{2} + 25} d x

\int e^{- r^{2}} r d r

\int_{0}^{\frac{π}{8}} 3 cos^{5} (4 x) d x

n = 0 \sum \infty \frac{x ^{2 n}}{2 n !}

\int 2 + 60 x - 5 x^{2} d x

What is the solution for y^'= x/y+(2y)/x ?
The solution for y^'= x/y+(2y)/x is y=xsqrt(-1+c_{1)x^2},y=-xsqrt(-1+c_{1)x^2}