What is the solution for (9y^2-t^2)/(y^5)(dy)/(dt)+t/(2y^4)=0 ?

The solution for (9y^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 >

(9y^2-t^2)/(y^5)(dy)/(dt)+t/(2y^4)=0

\frac{9 y ^{2} - t ^{2}}{y ^{5}} \frac{d y}{d t} + \frac{t}{2 y ^{4}} = 0

y = \frac{- 9 + 2 c _{1} t ^{2} + 81}{2 c _{1}}, y = - \frac{- 9 + 2 c _{1} t ^{2} + 81}{2 c _{1}}, y = - \frac{2 c _{1} t ^{2} + 81 + 9}{2 c _{1}}, y = - - \frac{2 c _{1} t ^{2} + 81 + 9}{2 c _{1}}

Solution steps

\frac{9 y ^{2} - t ^{2}}{y ^{5}} \frac{d y}{d t} + \frac{t}{2 y ^{4}} = 0

Solve exact ODE:

y = \frac{- 9 + 2 c _{1} t ^{2} + 81}{2 c _{1}}, y = - \frac{- 9 + 2 c _{1} t ^{2} + 81}{2 c _{1}}, y = - \frac{2 c _{1} t ^{2} + 81 + 9}{2 c _{1}}, y = - - \frac{2 c _{1} t ^{2} + 81 + 9}{2 c _{1}}

y = \frac{- 9 + 2 c _{1} t ^{2} + 81}{2 c _{1}}, y = - \frac{- 9 + 2 c _{1} t ^{2} + 81}{2 c _{1}}, y = - \frac{2 c _{1} t ^{2} + 81 + 9}{2 c _{1}}, y = - - \frac{2 c _{1} t ^{2} + 81 + 9}{2 c _{1}}

\frac{d y}{d x} - y = e^{x} \cdot y^{2}

\int_{4}^{\infty} 5 e^{- \frac{y}{2}} d y

\int \frac{x}{x + 2 y} d x

\frac{x ^{2} - 1}{x ^{3}}

\frac{d v}{d t} = 2.1 - \frac{7 v}{2180}, v (0) = 0

What is the solution for (9y^2-t^2)/(y^5)(dy)/(dt)+t/(2y^4)=0 ?
The solution for (9y^2-t^2)/(y^5)(dy)/(dt)+t/(2y^4)=0 is