initial value problem,

Question

Solve the initial value problem.

$\frac{dy}{dx} = y - y^2, y(0)=-\frac13$ Ans: $-\frac{4y}{1-y}=e^x$

I thought first to use integrating factor then I give it up since it cannot be converted into standart first order diff. eq. form. I cannot get the perspective to solve this type of question. Thanks for any help!

Answer 1

Rewrite the ode as

$$\frac{dy}{y-y^2} = dx$$

or,

$$\frac{dy}{y}+\frac{dy}{1-y}= dx$$

Integrate both sides with the initial value $y(0)=-1/3$,

$$[\ln (-y) - \ln(1-y)]-[\ln (-(-1/3)) - \ln(1-(-1/3))]=x$$

Simplify, $$\ln\left(\frac{-y}{1-y}\right)+\ln 4= x$$

to arrive at the result,

$$-\frac{4y}{1-y} = e^x$$

Answer 2

$\frac{dy}{dx} = y - y^2$ is a separable differential equation

multiply each side by dx and divide by the RHS

$\frac{dy}{y - y^2} = dx$ can be directly integrated and then you can apply the initial conditions