en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

Solution

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

Solution

y = c_{1} 1 + x^{2}

Solution steps

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

Solve separable ODE:

y = c_{1} 1 + x^{2}

y = c_{1} 1 + x^{2}

Popular Examples

2xtan(y)dx+sec^2(y)dy=0

2 x tan (y) d x + sec^{2} (y) d y = 0

u^'=csc(\H(x))cos(x/2)

u^{^{'}} = csc (H (x)) cos (\frac{x}{2})

y^'=e^xcos^2(y),y(0)= pi/4

y^{^{'}} = e^{x} cos^{2} (y), y (0) = \frac{π}{4}

(dy)/(dx)=-3/2 y+1/2 y^2

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

a d a + b d b = 0

Frequently Asked Questions (FAQ)

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