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Popular >

y^'=ky^2

Solution

y^{^{'}} = k y^{2}

Solution

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

Solution steps

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

Solve separable ODE:

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

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

Popular Examples

yy^'=(y^2-9)/(x^2+4)

y y^{^{'}} = \frac{y ^{2} - 9}{x ^{2} + 4}

(dy}{dx}=-\frac{2sin(x))/y

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

y(dy)/(dx)=3x(y^2+5)^5

y \frac{d y}{d x} = 3 x (y^{2} + 5)^{5}

dy=(y+9)/(x+8)dx

d y = \frac{y + 9}{x + 8} d x

cos^3(y)(dy)/(dx)=xcsc(y),y(0)= pi/3

cos^{3} (y) \frac{d y}{d x} = x csc (y), y (0) = \frac{π}{3}

Frequently Asked Questions (FAQ)

What is the solution for y^'=ky^2 ?
The solution for y^'=ky^2 is y=-1/(kt+c_{1)}