What is the solution for (x^2}{y^2-1}y^'=\frac{3x^3)/y ,y(0)=2 ?

The solution for (x^2}{y^2-1}y^'=\frac{3x^3)/y ,y(0)=2 is y=sqrt(3e^{3x^2)+1}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

(x^2}{y^2-1}y^'=\frac{3x^3)/y ,y(0)=2

\frac{x ^{2}}{y ^{2} - 1} y^{^{'}} = \frac{3 x ^{3}}{y}, y (0) = 2

y = 3 e^{3 x^{2}} + 1

Solution steps

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

Solve separable ODE:

y = 3 e^{3 x^{2}} + 1

y = 3 e^{3 x^{2}} + 1

What is the solution for (x^2}{y^2-1}y^'=\frac{3x^3)/y ,y(0)=2 ?
The solution for (x^2}{y^2-1}y^'=\frac{3x^3)/y ,y(0)=2 is y=sqrt(3e^{3x^2)+1}