What is the solution for (x^2)/(y^2-2)y^'= 1/(2y),y(1)=sqrt(3) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(x^2)/(y^2-2)y^'= 1/(2y),y(1)=sqrt(3)

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

y = e^{- \frac{1}{x} + 1} + 2

Solution steps

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

Solve separable ODE:

y = e^{- \frac{1}{x} + 1} + 2

y = e^{- \frac{1}{x} + 1} + 2

\int \frac{2 x ^{4} + 3 x ^{2} - 2 x ^{- 3}}{x ^{2}} d x

n = 1 \sum \infty \frac{1}{6 n + 5}

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

\int e^{3 x} sin (3 x) d x

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

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