Cant solve this differential equation

  • Thread starter pakmingki
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Homework Statement


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


Homework Equations


none really


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.
 

Answers and Replies

  • #2
Dick
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There's a sign error on the first term in your solution.
 
  • #3
benorin
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You solved the given differential equation [itex]\frac{dy}{dx}=x(y-1)^2[/itex] and obtained the correct general solution (except you dropped a - sign on the [itex]x^2[/itex] term after you multiplied by -1/2) [tex]y_g=\frac{2}{2G-x^2}+1[/tex] for which [tex]\frac{dy_g}{dx}=\frac{4x}{(2G-x^2)^2}=x(y_g-1)^2[/tex] 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 [itex]y\neq 1[/itex] 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 [tex]y_g=\frac{2}{2G-x^2}+1[/tex] we get [tex]1=\frac{1}{G}+1\Rightarrow \frac{1}{G}=0[/tex] which only holds if we let [itex]G\to\pm\infty[/itex] and this gives the particular solution [tex]y_p=\lim_{G\to\pm\infty}\left(\frac{2}{2G-x^2}+1\right)=0+1=1[/tex], which is the equilibrium solution.
 
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