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Popular >

y^'-y=-y^2

Solution

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

Solution

y = - \frac{e ^{t + c_{1}}}{1 - e ^{t + c_{1}}}

Solution steps

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

Solve separable ODE:

y = - \frac{e ^{t + c_{1}}}{1 - e ^{t + c_{1}}}

y = - \frac{e ^{t + c_{1}}}{1 - e ^{t + c_{1}}}

Popular Examples

5x+yy^'=0,y(2)=5

5 x + y y^{^{'}} = 0, y (2) = 5

(dy)/(dx)=x^2(-6y+9)

\frac{d y}{d x} = x^{2} (- 6 y + 9)

x^3(dy)/(dx)=sec(y)

x^{3} \frac{d y}{d x} = sec (y)

(dy)/(dx)+(e^{x/2})/(25e^{5y)}=0

\frac{d y}{d x} + \frac{e ^{\frac{x}{2}}}{25 e ^{5 y}} = 0

(t^2-4)y^'+ty=0

(t^{2} - 4) y^{^{'}} + t y = 0

Frequently Asked Questions (FAQ)

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