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Popular Calculus >

2y^'=y^2,y(0)=7

Solution

2 y^{^{'}} = y^{2}, y (0) = 7

Solution

y = - \frac{14}{7 t - 2}

Solution steps

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

Solve separable ODE:

y = - \frac{14}{7 t - 2}

y = - \frac{14}{7 t - 2}

Graph

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\int \frac{x ^{4}}{x ^{5} - 3} d x

sum from n=3 to infinity of (7/11)^{-n}

n = 3 \sum \infty (\frac{7}{11})^{- n}

Frequently Asked Questions (FAQ)

What is the solution for 2y^'=y^2,y(0)=7 ?
The solution for 2y^'=y^2,y(0)=7 is y=-(14)/(7t-2)