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Popular >

y^'=(x+xy)/(1+x^2)

Solution

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

Solution

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

Solution steps

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

Solve separable ODE:

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

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

Popular Examples

(dy)/(dx)x^2+2x^2y+3xy=0

\frac{d y}{d x} x^{2} + 2 x^{2} y + 3 x y = 0

(dx)/(dt)=20-0.4x

\frac{d x}{d t} = 20 - 0.4 x

(1 - y^{2}) d x = x y d y

1/y (dy)/(dx)=x^{-1}+x^2

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

(dy)/(dx)=y^2+1,y(0)=0

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

Frequently Asked Questions (FAQ)

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