Question

using separation of variables.. how can you get the solution of this Differential equation. (1-x)dy=y^2

Answer 1

The first thing you have to do is make sure all variables with y in them are on one side and all variables with x are on the other side

So once you move everything around, it should look like this

\[1/(y^2) dy = 1/(1-x)\]

And then we take the integral of both sides

Which is

\[\int\limits_{}^{}(\frac{ 1 }{ y^2 })dy = \int\limits_{}^{}\frac{ 1 }{ 1-x } dx\]

This gives us

\[\frac{ -1 }{ y} = -\ln (1-x) + c\]

Where c is a constant of integration. If you multiply by y and divide by the stuff on the right you get:

\[y = \frac{ -1 }{ -\ln(1-x) +c }\]

Which is a more conventional form of writing equations. If you have values for x and y, you can plug them in to find c. Otherwise, this is the form of the general solution.

Answer 2

oops wait.. i am confused to ur solution.. how come that dx exist when you integrate 1/(1-x) even it doesnt have dx on the given..

Answer 3

Hmm there's no dx to start with? Just the differential dy..?
Is it by change a big D?
Dy

Answer 4

what will happen to 1/(1-x) ? is it possible to integrate even without dx??

Answer 5

No, I don't think we can. That's why I'm trying to understand if the problem was pasted correctly c: Seems like there might be a small typo somewhere.

Answer 6

hmmm??? i think your right.. but i dont have chance to see if there's a typo error on the problem.. by the way.. thanks..

Answer 7

what if you have \[ dy = 1/(1-x)* 1/(y^2)\] then
\[\int\limits_{}^{} dy = \frac{ 1 }{ (1-x)y ^{2} }\]
\[y+c = \frac{ 1 }{ (1-x)y ^{2} }\]
\[y ^{2}(y+c) = \frac{ 1 }{ (1-x) }\]
\[y ^{3}+cy^{2} = \frac{ 1 }{ (1-x) }\]
does this make any sense?