another differential

Dec 2008
105
3
Hi, again

I am slightly confused by what the questions is asking here.

Solve the differential to find y

\(\displaystyle \frac{dy}{dx} = xy(1 - y)\) subject to y = 0.5 when x = 0. What is the purpose of the 'subject....' part?
 
May 2010
43
1
The "Subject to...." is giving you the boundary conditions needed to solve the problem.
 
Dec 2008
105
3
What is the point working it out if you already have the y value, is it just to get another y value at the end???
 

HallsofIvy

MHF Helper
Apr 2005
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7,909
You don't have "the y value". y is a function of x. Telling you that "y(0)= 0.5" only tells you its value for one value of x. You want to find the entire function.

This is a "separable" equation. If \(\displaystyle \frac{dy}{dx}= xy(1- y)\) then \(\displaystyle \frac{dy}{y(1-y)}= xdx\). Integrate both sides of that (use "partial fractions" on the left) to get y as a function of x. When you integrate, you will get a "constant of integration". You can use y(0)= .5 to determine what that constant should be.
 
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May 2010
43
1
BTW, when I solve this I obtain \(\displaystyle y/(1-y) = C*e^{x^2/2}\). The boundary condition y=0.5 at x=0 gives C=1.
 
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Prove It

MHF Helper
Aug 2008
12,883
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What is the point working it out if you already have the y value, is it just to get another y value at the end???
It's because when you integrate a function, the solution is a FAMILY of functions, due to there being an integration constant. The point of having a boundary condition is that you can solve for the integration constant.

In your case:

\(\displaystyle \frac{dy}{dx} = x\,y\,(1 - y)\)

\(\displaystyle \frac{1}{y\,(1 - y)}\,\frac{dy}{dx} = x\).


Using partial fractions:

\(\displaystyle \frac{A}{y} + \frac{B}{1 - y} = \frac{1}{y\,(1 - y)}\)

\(\displaystyle \frac{A(1 - y) + B\,y}{y\,(1 - y)} = \frac{1}{y\,(1 - y)}\)

\(\displaystyle A(1 - y) + B\,y = 1\)

\(\displaystyle (B - A)\,y + A = 0\,y + 1\).

Equating like coefficients gives

\(\displaystyle B - A = 0\) and \(\displaystyle A = 1\).

So \(\displaystyle B = 1\).

Therefore \(\displaystyle \frac{1}{y\,(1 - y)} = \frac{1}{y} + \frac{1}{1 - y}\).


Back to the DE:

\(\displaystyle \frac{1}{y\,(1 - y)}\,\frac{dy}{dx} = x\)

\(\displaystyle \left(\frac{1}{y} + \frac{1}{1 - y}\right)\,\frac{dy}{dx} = x\)

\(\displaystyle \int{\left(\frac{1}{y} + \frac{1}{1 - y}\right)\,\frac{dy}{dx}\,dx} = \int{x\,dx}\)

\(\displaystyle \int{\left(\frac{1}{y} + \frac{1}{1 - y}\right)\,dy} = \frac{1}{2}\,x^2 + C_1\)

\(\displaystyle \ln{|y|} - \ln{|1 - y|} + C_2 = \frac{1}{2}\,x^2 + C_1\)

\(\displaystyle \ln{\left|\frac{y}{1 - y}\right|} = \frac{1}{2}\,x^2 + C\) where \(\displaystyle C = C_1 - C_2\)

\(\displaystyle \ln{\left|\frac{y - 1 + 1}{1 - y}\right|} = \frac{1}{2}\,x^2 + C\)

\(\displaystyle \ln{\left|-\left(\frac{1 - y}{1 - y}\right) + \frac{1}{1 - y}\right|} = \frac{1}{2}\,x^2 + C\)

\(\displaystyle \ln{\left|-1 + \frac{1}{1 - y}\right|} = \frac{1}{2}\,x^2 + C\)

\(\displaystyle \left|-1 + \frac{1}{1 - y}\right| = e^{\frac{1}{2}\,x^2 + C}\)

\(\displaystyle \left|-1 + \frac{1}{1 - y}\right| = e^C\,e^{\frac{1}{2}\,x^2}\)

\(\displaystyle -1 + \frac{1}{1 - y} = A\,e^{\frac{1}{2}\,x^2}\) where \(\displaystyle A = \pm e^{C}\)

\(\displaystyle \frac{1}{1 - y} = A\,e^{\frac{1}{2}\,x^2} + 1\)

\(\displaystyle 1 - y = \frac{1}{A\,e^{\frac{1}{2}\,x^2}}\)

\(\displaystyle y = 1 - \frac{1}{A\,e^{\frac{1}{2}\,x^2}}\).


Now using the boundary condition that \(\displaystyle y(0) = \frac{1}{2}\) we find

\(\displaystyle \frac{1}{2} = 1 - \frac{1}{A\,e^{\frac{1}{2}(0)^2}}\)

\(\displaystyle \frac{1}{A\,e^{0}} = 1 - \frac{1}{2}\)

\(\displaystyle \frac{1}{A} = \frac{1}{2}\)

\(\displaystyle A = 2\).


So finally, the solution is

\(\displaystyle y = 1 - \frac{1}{2\,e^{\frac{1}{2}\,x^2}}\)

\(\displaystyle y = \frac{2\,e^{\frac{1}{2}\,x^2} - 1}{2\,e^{\frac{1}{2}\,x^2}}\).
 
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HallsofIvy

MHF Helper
Apr 2005
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7,909
Oh, you shouldn't tell him that! Let Beard work it out himself.(Wink)
 

Prove It

MHF Helper
Aug 2008
12,883
4,999
Oh, you shouldn't tell him that! Let Beard work it out himself.(Wink)
Judging by the OP's statements and questions, it appears that he/she hasn't had much practice with these sorts of problems. So by having a full solution, at least the OP has something to refer back to for future similar problems.
 
Dec 2008
105
3
Oh, you shouldn't tell him that! Let Beard work it out himself.(Wink)

I had worked the integrals out before I posted its just that I was unsure of why/how the boundary conditions were used