What is the solution for ((3y^2-t^2))/(y^5)(dy)/(dt)+t/(2y^4)=0 ?

The solution for ((3y^2-t^2))/(y^5)(dy)/(dt)+t/(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-t^2))/(y^5)(dy)/(dt)+t/(2y^4)=0

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

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

Solution steps

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

Solve exact ODE:

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

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

\frac{\partial}{\partial y} (cos (x^{3} y^{3}))

\frac{\partial}{\partial x} (- 2 x y sin (x y^{2}))

tan^{2} (x)

y^{^{'}} = y - x^{2} + 1, y (0) = 0

\int \frac{b - x}{x} d x

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