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Popular >

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

Solution

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

Solution

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

Solution steps

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

Solve separable ODE:

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

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

Popular Examples

derivative of (4x+1^2(1-x)^3)

\frac{d}{d x} ((4 x + 1)^{2} (1 - x)^{3})

limit as x approaches 0 of (x+1)/(x+4)

x \to 0 lim (\frac{x + 1}{x + 4})

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

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

sqrt(y^2+1)dx=xydy

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

derivative of x/(x^2+t^2)

\frac{d}{d x} (\frac{x}{x ^{2} + t ^{2}})

Frequently Asked Questions (FAQ)

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