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Popular >

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

Solution

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

Solution

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

Solution steps

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

Solve separable ODE:

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

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

Popular Examples

2xy+6x+(x^2-4)y^'=0,y(3)=2

2 x y + 6 x + (x^{2} - 4) y^{^{'}} = 0, y (3) = 2

(dy)/(dx)=-(0.6)/(cos(y)),y(0)= pi/4

\frac{d y}{d x} = - \frac{0.6}{cos ( y )}, y (0) = \frac{π}{4}

(dy)/(dx)=x^5e^{-y}

\frac{d y}{d x} = x^{5} e^{- y}

2(x+1)yy^'-y^2=2,y(5)=2

2 (x + 1) y y^{^{'}} - y^{2} = 2, y (5) = 2

2sin(y^2)dx+xycos(y^2)dy=0

2 sin (y^{2}) d x + x y cos (y^{2}) d y = 0

Frequently Asked Questions (FAQ)

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