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Popular >

y^'-y^2=0,y(0)=1

Solution

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

Solution

y = - \frac{1}{t - 1}

Solution steps

y^{'} (t) - y^{2} = 0

Solve separable ODE:

y = - \frac{1}{t - 1}

y = - \frac{1}{t - 1}

Popular Examples

integral from 0 to 1 of (x^8)/(x^9-1)

\int_{0}^{1} \frac{x ^{8}}{x ^{9} - 1} d x

integral of x(x^2)

\int x (x^{2}) d x

integral from 1 to 2 of-2x+6

\int_{1}^{2} - 2 x + 6 d x

derivative of 1/(x^{16})

\frac{d}{d x} (\frac{1}{x ^{16}})

area y=-8x,y=12,x=-1,x=1 area

y = - 8 x, y = 12, x = - 1, x = 1

Frequently Asked Questions (FAQ)

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