dy/dx = (y- y^2)/ x for all x not zero; verify that y = ....

xc630

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dy/dx = (y- y^2)/ x for all x is not equal to 0.

Verify that y= x/(x+C), x is not equal to -C is a general solution for the given differential equation and show that all solutions contain ( 0,0).

I am not sure what I am being asked to do. How owuld i find a solution for equation? Set it equal to zero? And where would the C have come from? Thanks.
 
I tried separating the variables and got x/ dx = (y-y^2)/ dy. the dx and dy are on teh bottom. CAn I get them to the top somehow?
 
You must've separated wrong then.

Try it this way:

\(\displaystyle \L\\\int\frac{dy}{y-y^{2}}=\int\frac{dx}{x}\)
 
Okay I still getting mixed up.

I have ln(x) + C= ln (y-y^2) + C
I fyou use the e power wouldn't you get x + C = y-y^2 + C. What did I do wrong because I do not think you can derive y= x/ (x+ C) from that.
 
\(\displaystyle \L\\\int\frac{dy}{y-y^{2}}=\int\frac{dx}{x}\)

\(\displaystyle \L\\ln\left(\frac{y}{y-1}\right)=ln(x)+c\)

\(\displaystyle \L\\\frac{y}{y-1}=e^{c}x\)

\(\displaystyle \L\\\frac{y}{y-1}=\overbrace{C}^{\text{e^c=constant}}x\)

\(\displaystyle \L\\y=\frac{Cx}{Cx-1}\)...[1]

This is a correct solution. How do they get \(\displaystyle \frac{x}{x+C}\)?.

No doubt, some little thing I am overlooking.

Sub [1] into the original equation to check.

\(\displaystyle \L\\\frac{d}{dx}\left[\frac{Cx}{Cx-1}\right]=\frac{-C}{(Cx-1)^{2}}\)

\(\displaystyle \L\\\frac{\left(\frac{Cx}{Cx-1}\right)-\left(\frac{Cx}{Cx-1}\right)^{2}}{x}=\frac{-C}{(Cx-1)^{2}}\)...check.
 
Galact, your C doesn't have to be their C. Call yours A, and divide top and bottom of your solution by A.

As for the original question, usually verifying a solution to a differential equation involves just differentiating and substituting in.
 
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