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

The solution for y^'=(x^2+2y^2)/(xy) 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

Popular >

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

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

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

Solution steps

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

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}

y^{^{''}} - y^{^{'}} - 30 = 0

569 + t (37 t^{0.6} - 108)

\frac{d r}{d θ} = \frac{r ^{2} + 10 r θ}{5 θ ^{2}}

\int 5 e^{2 x - 1} d x

y^{^{''}} - 18 y^{^{'}} + 77 y = 0

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