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

The solution for ((3y^2-t^2))/(y^5)dy+t/(2y^4)dt=0 is t=ysqrt(6+c_{1)y^2},t=-ysqrt(6+c_{1)y^2}

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+t/(2y^4)dt=0

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

t = y 6 + c_{1} y^{2}, t = - y 6 + c_{1} y^{2}

Solution steps

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

Solve Bernoulli ODE:

t = y 6 + c_{1} y^{2}, t = - y 6 + c_{1} y^{2}

t = y 6 + c_{1} y^{2}, t = - y 6 + c_{1} y^{2}

n = 0 \sum \infty (\frac{3}{5})^{n}

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

\int x \cdot 2^{x} d x

\int 4 cos (4 x) d x

y = cos^{7} (π t - 16)

What is the solution for ((3y^2-t^2))/(y^5)dy+t/(2y^4)dt=0 ?
The solution for ((3y^2-t^2))/(y^5)dy+t/(2y^4)dt=0 is t=ysqrt(6+c_{1)y^2},t=-ysqrt(6+c_{1)y^2}