What is the solution for y^'=2,xy^2y= 1/((4-x^2)) ?

The solution for y^'=2,xy^2y= 1/((4-x^2)) is y=2x+c_{1}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'=2,xy^2y= 1/((4-x^2))

y^{^{'}} = 2, x y^{2} y = \frac{1}{( 4 - x ^{2} )}

y = 2 x + c_{1}

Solution steps

y^{'} (x) = 2

f^{^{'}} (x) = g (x)

then

f (x) = \int g (x) d x

y = \int 2 d x

\int 2 d x = 2 x + c_{1}

y = 2 x + c_{1}

f^{^{'}} (x) = x^{3} + 8 x - 5 x + 6

y^{^{'}} = \frac{( 2 )}{( x ^{3} ) + 5 x ^{\frac{( 2 )}{( 3 )}}}

x^{2} y^{^{'}} = 1

f^{^{'}} (x) = 9 x^{2} + 10 x

\frac{y ^{^{'}} + 1}{x} = \frac{e ^{x}}{x ^{2}}

What is the solution for y^'=2,xy^2y= 1/((4-x^2)) ?
The solution for y^'=2,xy^2y= 1/((4-x^2)) is y=2x+c_{1}