# Cant solve this differential equation

1. Feb 26, 2007

### pakmingki

1. The problem statement, all variables and given/known data
a. find the general solution to this differential equation
dy/dx = x(y-1)^2
b. find the particular solution to the given initial condition f(0) = 1
c. use the solution found in b to find the range of f

2. Relevant equations
none really

3. The attempt at a solution
this question seemed simple, but i cant really get the right answer.
here is my attempt

*i use different letters after manipulating constants, because they are still constants after adding/subtracting/dividing/multiplying

dy/dx = x(y-1)^2

*separation of variables*

x dx = (y-1)^-2 dy

*integrate both sides*

x^2/2 + C1= -(y-1)^-1 + C2

x^2/2 = -(y-1)^-1 + K

solve for y

(x^2/2 = -1/(y-1) + K) * 2

x^2 = -2/(y-1) + L

x^2 = -2(y-1)^-1 + L

(x^2 + H = -2(y-1)^-1 ) * -1/2
x^2/2 + G = (y-1)^-1
raise everything to the negative 1

(x^2/2 +G)^-1 = y-1

x^2/2 +G = (x^2 + 2G)/2

((x^2 + 2G)/2)^-1 = 2/(x^2 + 2G)

2/(x^2 + 2G) + 1 = y

but i can tell already this is the wrong answer, and with a wrong general solution, i cant do the other parts of the problem.

2. Feb 26, 2007

### Dick

There's a sign error on the first term in your solution.

3. Feb 26, 2007

### benorin

You solved the given differential equation $\frac{dy}{dx}=x(y-1)^2$ and obtained the correct general solution (except you dropped a - sign on the $x^2$ term after you multiplied by -1/2) $$y_g=\frac{2}{2G-x^2}+1$$ for which $$\frac{dy_g}{dx}=\frac{4x}{(2G-x^2)^2}=x(y_g-1)^2$$ and is then indeed a valid solution to the given equation, however the initial condition that f(0)=1 is something of a problem since when we solved the equation we had divided both sides by (y-1)^2 for which we need to assume that $y\neq 1$ since that would involve division by zero. But maybe we can ignore that and solve it anyways... If we set x=0 and y=1 in the general solution $$y_g=\frac{2}{2G-x^2}+1$$ we get $$1=\frac{1}{G}+1\Rightarrow \frac{1}{G}=0$$ which only holds if we let $G\to\pm\infty$ and this gives the particular solution $$y_p=\lim_{G\to\pm\infty}\left(\frac{2}{2G-x^2}+1\right)=0+1=1$$, which is the equilibrium solution.

Last edited: Feb 26, 2007