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Popular >

y^'=1+y^2

Solution

y^{^{'}} = 1 + y^{2}

Solution

y = tan (t + c_{1})

Solution steps

y^{'} (t) = 1 + y^{2}

Solve separable ODE:

y = tan (t + c_{1})

y = tan (t + c_{1})

Popular Examples

derivative of f(x)=3x^2-x+3 derivative of

f (x) = 3 x^{2} - x + 3

derivative of (x-17/(sqrt(x)-\sqrt{17)})

\frac{d}{d x} (\frac{x - 17}{x - 17})

integral of sqrt(4-x^2)(16-4x^2)

\int 4 - x^{2} (16 - 4 x^{2}) d x

integral of cos(xz)

\int cos (x z) d x

(dy)/(dx)=0.0012y(2300-y)

\frac{d y}{d x} = 0.0012 y (2300 - y)

Frequently Asked Questions (FAQ)

What is the solution for y^'=1+y^2 ?
The solution for y^'=1+y^2 is y=tan(t+c_{1})