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Popular >

(1+y^2)dx-dy=0

Solution

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

Solution

y = tan (x + c_{1})

Solution steps

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

Solve separable ODE:

y = tan (x + c_{1})

y = tan (x + c_{1})

Popular Examples

y^'=-y+1,y(0)=0

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

(dy)/(dx)=ye^{-x^2},y(5)=1

\frac{d y}{d x} = y e^{- x^{2}}, y (5) = 1

1/2 yy^'=t^3,y(0)=2

\frac{1}{2} y y^{^{'}} = t^{3}, y (0) = 2

(x^2+9)dy=(2x-8xy)dx

(x^{2} + 9) d y = (2 x - 8 x y) d x

y^'=(2sin(x))/(sin(y)),y(0)= pi/2

y^{^{'}} = \frac{2 sin ( x )}{sin ( y )}, y (0) = \frac{π}{2}

Frequently Asked Questions (FAQ)

What is the solution for (1+y^2)dx-dy=0 ?
The solution for (1+y^2)dx-dy=0 is y=tan(x+c_{1})