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Popular >

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

Solution

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

Solution

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

Solution steps

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

Solve separable ODE:

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

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

Graph

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Frequently Asked Questions (FAQ)

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