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Popular >

y^'=1+4y^2,y(0)=0

Solution

y^{^{'}} = 1 + 4 y^{2}, y (0) = 0

Solution

y = \frac{tan ( 2 t )}{2}

Solution steps

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

Solve separable ODE:

y = \frac{tan ( 2 t )}{2}

y = \frac{tan ( 2 t )}{2}

Graph

Popular Examples

y^'=((e^{-x}-e^x))/((3+4y))

y^{^{'}} = \frac{( e ^{- x} - e ^{x} )}{( 3 + 4 y )}

y^{^{'}} = y (1 - y^{2})

9ty+e^ty^'= y/(t^2+81)

9 t y + e^{t} y^{^{'}} = \frac{y}{t ^{2} + 81}

x^{^{'}} - e^{t} x = 5 e^{t}

y^'+2xy=x,y(0)= 1/2+1/2 e^{-x^2}

y^{^{'}} + 2 x y = x, y (0) = \frac{1}{2} + \frac{1}{2} e^{- x^{2}}

Frequently Asked Questions (FAQ)

What is the solution for y^'=1+4y^2,y(0)=0 ?
The solution for y^'=1+4y^2,y(0)=0 is y=(tan(2t))/2