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

The solution for x^2y^'=y^2+x^2 is y=(sqrt(3)xtan((sqrt(3)(ln(x)+c_{1}))/2))/(2)+1/2 x

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

x^2y^'=y^2+x^2

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

y = \frac{3 x tan ( \frac{3 ( l n ( x ) + c _{1} )}{2} )}{2} + \frac{1}{2} x

Solution steps

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

Solve using substitution:

y = \frac{3 x tan ( \frac{3 ( l n ( x ) + c _{1} )}{2} )}{2} + \frac{1}{2} x

y = \frac{3 x tan ( \frac{3 ( l n ( x ) + c _{1} )}{2} )}{2} + \frac{1}{2} x

\frac{d}{d x} ((x + 2) e^{- x})

\frac{d y}{d x} = \frac{x}{34 + x ^{2}}

\frac{d y}{d x} = 9 y e^{x + 3}

\int \frac{3 + 16 x}{8 x - 4} d x

\frac{d}{d x} (- \frac{x}{4} + \frac{3}{4})

What is the solution for x^2y^'=y^2+x^2 ?
The solution for x^2y^'=y^2+x^2 is y=(sqrt(3)xtan((sqrt(3)(ln(x)+c_{1}))/2))/(2)+1/2 x