What is the solution for ((3y^2-x^2))/(y^5)y^'+x/(2y^4)=0 ?

The solution for ((3y^2-x^2))/(y^5)y^'+x/(2y^4)=0 is

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

((3y^2-x^2))/(y^5)y^'+x/(2y^4)=0

\frac{( 3 y ^{2} - x ^{2} )}{y ^{5}} y^{^{'}} + \frac{x}{2 y ^{4}} = 0

y = \frac{- 3 + 2 c _{1} x ^{2} + 9}{2 c _{1}}, y = - \frac{- 3 + 2 c _{1} x ^{2} + 9}{2 c _{1}}, y = - \frac{2 c _{1} x ^{2} + 9 + 3}{2 c _{1}}, y = - - \frac{2 c _{1} x ^{2} + 9 + 3}{2 c _{1}}

Solution steps

\frac{( 3 y ^{2} - x ^{2} )}{y ^{5}} y^{'} (x) + \frac{x}{2 y ^{4}} = 0

Solve exact ODE:

y = \frac{- 3 + 2 c _{1} x ^{2} + 9}{2 c _{1}}, y = - \frac{- 3 + 2 c _{1} x ^{2} + 9}{2 c _{1}}, y = - \frac{2 c _{1} x ^{2} + 9 + 3}{2 c _{1}}, y = - - \frac{2 c _{1} x ^{2} + 9 + 3}{2 c _{1}}

y = \frac{- 3 + 2 c _{1} x ^{2} + 9}{2 c _{1}}, y = - \frac{- 3 + 2 c _{1} x ^{2} + 9}{2 c _{1}}, y = - \frac{2 c _{1} x ^{2} + 9 + 3}{2 c _{1}}, y = - - \frac{2 c _{1} x ^{2} + 9 + 3}{2 c _{1}}

f (x) = - 5.7 + 3.57 ln (x)

\int 1 + cos^{2} (x) d x

\frac{d}{d x} (\frac{3 x - 1}{x + 1})

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

\frac{d}{d x} (8 (0.9)^{x - 2})

What is the solution for ((3y^2-x^2))/(y^5)y^'+x/(2y^4)=0 ?
The solution for ((3y^2-x^2))/(y^5)y^'+x/(2y^4)=0 is