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Popular >

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

Solution

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

Solution

y = - \frac{1}{arctan ( x ) + c _{1}}

Solution steps

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

Solve separable ODE:

y = - \frac{1}{arctan ( x ) + c _{1}}

y = - \frac{1}{arctan ( x ) + c _{1}}

Popular Examples

y^'=(e^x)/(e^y),y(0)=ln(2)

y^{^{'}} = \frac{e ^{x}}{e ^{y}}, y (0) = ln (2)

(dy)/(dx)=-4/6 x^4y^8

\frac{d y}{d x} = - \frac{4}{6} x^{4} y^{8}

(dy}{dx}=\frac{sqrt(y))/x

\frac{d y}{d x} = \frac{y}{x}

d v = x d x

y^'=(-0.8)/(cos(y)),y(0)= pi/2

y^{^{'}} = \frac{- 0.8}{cos ( y )}, y (0) = \frac{π}{2}

Frequently Asked Questions (FAQ)

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