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Popular >

y^'=-y^2,y(0)=0.5

Solution

y^{^{'}} = - y^{2}, y (0) = 0.5

Solution

y = \frac{1}{t + 2}

Solution steps

y^{'} (t) = - y^{2}

Solve separable ODE:

y = \frac{1}{t + 2}

y = \frac{1}{t + 2}

Graph

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Frequently Asked Questions (FAQ)

What is the solution for y^'=-y^2,y(0)=0.5 ?
The solution for y^'=-y^2,y(0)=0.5 is y= 1/(t+2)