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Popular >

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

Solution

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

Solution

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

Solution steps

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

Solve separable ODE:

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

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

Graph

Popular Examples

tangent of y=x^4+9x^2-x,(1,9)tangent of

y = x^{4} + 9 x^{2} - x, (1, 9)

e^{x + y} - y^{^{'}} = 0

(dy)/(dx)=(x-2ycot(2x))

\frac{d y}{d x} = (x - 2 y cot (2 x))

integral of (7sqrt(x))/(x^3-4)

\int \frac{7 x}{x ^{3} - 4} d x

integral of ((x-4)/(x^9))

\int (\frac{x - 4}{x ^{9}}) d x

Frequently Asked Questions (FAQ)

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