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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Solve using substitution:

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

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

\int_{2}^{5} \frac{x}{x - 1} d x

\frac{d}{d x} (\frac{1}{4 x ^{2} + 1})

\frac{\partial}{\partial x} (\frac{( 2 x + 3 y )}{4 x + 5 y})

f (x) = x^{2} + 5

cos (π x), 1

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