What is the solution for yy^'= x/(x^2+1),y(0)=-2 ?

The solution for yy^'= x/(x^2+1),y(0)=-2 is y=-sqrt(ln(x^2+1)+4)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

yy^'= x/(x^2+1),y(0)=-2

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

y = - ln (x^{2} + 1) + 4

Solution steps

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

Solve separable ODE:

y = - ln (x^{2} + 1) + 4

y = - ln (x^{2} + 1) + 4

\frac{d y}{d x} - \frac{y}{3 x} + \frac{x}{3 y} = 0

\int x sin^{2} (6 x) d x

\frac{d}{d x} (4 x^{3} - 46 x^{2} + 120 x)

\int \frac{1}{x} d x

\int sin (5 x) cos (x) d x

What is the solution for yy^'= x/(x^2+1),y(0)=-2 ?
The solution for yy^'= x/(x^2+1),y(0)=-2 is y=-sqrt(ln(x^2+1)+4)