What is the solution for (x^2}{y^2-1}y^'=\frac{3x^3)/y ?

The solution for (x^2}{y^2-1}y^'=\frac{3x^3)/y is y=sqrt(e^{3x^2+c_{1)}+1},y=-sqrt(e^{3x^2+c_{1)}+1}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(x^2}{y^2-1}y^'=\frac{3x^3)/y

\frac{x ^{2}}{y ^{2} - 1} y^{^{'}} = \frac{3 x ^{3}}{y}

y = e^{3 x^{2} + c_{1}} + 1, y = - e^{3 x^{2} + c_{1}} + 1

Solution steps

\frac{x ^{2}}{y ^{2} - 1} y^{'} (x) = \frac{3 x ^{3}}{y}

Solve separable ODE:

y = e^{3 x^{2} + c_{1}} + 1, y = - e^{3 x^{2} + c_{1}} + 1

y = e^{3 x^{2} + c_{1}} + 1, y = - e^{3 x^{2} + c_{1}} + 1

y^{^{'}} = 9 y

(x + 1) \frac{d y}{d x} + y = 10

x^{^{'}} + x = - 1

y^{^{'}} = \frac{x y sin ( x ^{2} )}{y + 1}

\frac{d y}{d x} + y x^{2} = x^{2}

What is the solution for (x^2}{y^2-1}y^'=\frac{3x^3)/y ?
The solution for (x^2}{y^2-1}y^'=\frac{3x^3)/y is y=sqrt(e^{3x^2+c_{1)}+1},y=-sqrt(e^{3x^2+c_{1)}+1}