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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

y = tan (\frac{1}{2} ln (x) + c_{1}) x

Solution steps

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

Solve using substitution:

y = tan (\frac{1}{2} ln (x) + c_{1}) x

y = tan (\frac{1}{2} ln (x) + c_{1}) x

f (x) = 3 x^{3} + 2 x^{2} + 1, at x = - 1

x \to - 2 - lim (\frac{- 3}{x + 2})

x \to 0 lim (ln (\frac{7}{x ^{2}}))

\frac{d}{d x} (\frac{5}{3 x - 4})

\frac{d}{d x} (∣ x ∣)

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