en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2xy+(1+x^2)*y^'=0

Solution

2 x y + (1 + x^{2}) \cdot y^{^{'}} = 0

Solution

y = \frac{c _{1}}{1 + x ^{2}}

Solution steps

2 x y + (1 + x^{2}) \cdot y^{'} (x) = 0

Solve separable ODE:

y = \frac{c _{1}}{1 + x ^{2}}

y = \frac{c _{1}}{1 + x ^{2}}

Graph

Popular Examples

y^'=8tcos^2(y),y(1)=(3pi)/4

y^{^{'}} = 8 t cos^{2} (y), y (1) = \frac{3 π}{4}

derivative of (x^8/(y^5))=0

\frac{d}{d x} (\frac{x ^{8}}{y ^{5}}) = 0

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

y=(\sqrt[3]{(1+3x^2)^2})/x (dy)/(dx)

y = \frac{3 ( 1 + 3 x ^{2} ) ^{2}}{x} \frac{d y}{d x}

(dy)/(dt)y=sqrt(3+csc(t^6))

\frac{d y}{d t} y = 3 + csc (t^{6})

Frequently Asked Questions (FAQ)

What is the solution for 2xy+(1+x^2)*y^'=0 ?
The solution for 2xy+(1+x^2)*y^'=0 is y=(c_{1})/(1+x^2)