What is the solution for 2xyy^'-2x^2+1=0 ?

The solution for 2xyy^'-2x^2+1=0 is y=sqrt(x^2-ln(x)+c_{1)},y=-sqrt(x^2-ln(x)+c_{1)}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2xyy^'-2x^2+1=0

2 x y y^{^{'}} - 2 x^{2} + 1 = 0

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

Solution steps

2 x y y^{'} (x) - 2 x^{2} + 1 = 0

Solve separable ODE:

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

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

x y^{^{'}} - y ln (y) = 0

y^{^{'}} - 4 x y = 2 x

sin (x) cos^{2} (y) d x - cos (x) sin (y) d y = 0

\frac{d y}{d x} = \frac{7 ( 1 + x )}{x y}

\frac{\frac{d y}{d x}}{1 - y} = \frac{x ^{2}}{2}

What is the solution for 2xyy^'-2x^2+1=0 ?
The solution for 2xyy^'-2x^2+1=0 is y=sqrt(x^2-ln(x)+c_{1)},y=-sqrt(x^2-ln(x)+c_{1)}