Solve following Separable Differential Equation. Explain is there any constant solution?

Question

Find a solution of $xy' = y^2- y$ that passes through indicated points : $$a)\space (1,0), \space b) \space(0,0) \space c)\space (1/2 , 1/2) \space d)\space (2, 1/4)$$

Answer 1

Note : It's always important to show some effort. This is not a homework solving site. Below I'll just elaborate on a partial answer to leave you with work to do.

First of all, you have to find a general solution, which you will use to apply the initial values given (the points) in order to conclude to a specific solution via constant calculation. Starting off :

$$xy' = y^2 - y \Leftrightarrow y' = \frac{y^2-y}{x} \Rightarrow \int \frac{y'}{y^2-y}dx=\int\frac{1}{x}dx $$

$$ \Leftrightarrow $$

$$\ln|-y(x)+1| - \ln|y(x)| = \ln|x| + c_1 \Leftrightarrow \ln\bigg|\frac{-y(x)+1}{y(x)}\bigg| = \ln|x| + c_1 $$

$$\Rightarrow \dots \Rightarrow$$

$$y(x) = \frac{1}{c_1x+1}$$

I left the steps in the calculations in-between for you to handle and figure out. Now, for the part of the points, you just have to express them as $(x_0,y(x_0))$ and apply them as $y(x_0) = y_0$ to find different values for $c_1$ each time. Finally, regarding the constant solution, what conclusion can you figure out by the elaboration of the points-part of the question ?

Answer 2

Another way to solve it...

$$xy' = y^2- y$$ $$xy' +y= y^2$$ $$(xy)'= y^2$$ $$(xy)'= \frac {(xy)^2}{x^2}$$ $$\int \frac {d(xy)}{(xy)^2}= \int \frac {dx}{x^2}$$ $$ \frac {1}{(xy)}= \frac {1}{x}+K$$ $$\boxed{y=\frac 1 {Kx+1}}$$