What is the solution for 2xyy^'=3y^2+x^2 ?

The solution for 2xyy^'=3y^2+x^2 is y=xsqrt(-1+c_{1)x},y=-xsqrt(-1+c_{1)x}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

2xyy^'=3y^2+x^2

2 x y y^{^{'}} = 3 y^{2} + x^{2}

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

Solution steps

2 x y y^{'} (x) = 3 y^{2} + x^{2}

Solve Bernoulli ODE:

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

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

x \to 1 + lim (3 x - 1)

\int \frac{1}{x ^{3} - 1} d x

im pl i c i t \frac{d y}{d x}, x^{2} - y^{3} = 6

\int cos (x) + \frac{1}{3} sin (x) - \frac{1}{x ^{2}} d x

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

What is the solution for 2xyy^'=3y^2+x^2 ?
The solution for 2xyy^'=3y^2+x^2 is y=xsqrt(-1+c_{1)x},y=-xsqrt(-1+c_{1)x}