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Popular >

y^'=1+4y^2

Solution

y^{^{'}} = 1 + 4 y^{2}

Solution

y = \frac{tan ( 2 t + c _{1} )}{2}

Solution steps

y^{'} (t) = 1 + 4 y^{2}

Solve separable ODE:

y = \frac{tan ( 2 t + c _{1} )}{2}

y = \frac{tan ( 2 t + c _{1} )}{2}

Popular Examples

(\partial)/(\partial t)(e^{2t})

\frac{\partial}{\partial t} (e^{2 t})

tangent of f(x)=(6x)/(x+5),\at x=-3 tangent of

f (x) = \frac{6 x}{x + 5}, at x = - 3

d/(dt)(\sqrt[3]{t^3+s}+\sqrt[3]{s^3+t})

derivative of ln(2x^2+5)

\frac{d}{d x} (ln (2 x^{2} + 5))

integral of (1+x+x^2+x^3)

\int (1 + x + x^{2} + x^{3}) d x

Frequently Asked Questions (FAQ)

What is the solution for y^'=1+4y^2 ?
The solution for y^'=1+4y^2 is y=(tan(2t+c_{1}))/2