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Popular >

y^'-y^2=0

Solution

y^{^{'}} - y^{2} = 0

Solution

y = - \frac{1}{t + c _{1}}

Solution steps

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

Solve separable ODE:

y = - \frac{1}{t + c _{1}}

y = - \frac{1}{t + c _{1}}

Popular Examples

derivative of sin^3(cos(8x))derivative of

sin^{3} (cos (8 x))

integral of cos^2(2x)cos(2x)

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

(\partial)/(\partial x)(4+ln(xy-5))

\frac{\partial}{\partial x} (4 + ln (x y - 5))

integral of 4^k

\int 4^{k} d k

f(θ)=(sin(θ))/(1+cos(θ))

f (θ) = \frac{sin ( θ )}{1 + cos ( θ )}

Frequently Asked Questions (FAQ)

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