flip it on its head i.e. dx/dy and then it becomes bernoulli i think

[QUOTE=waqas izhar;747802]flip it on its head i.e. dx/dy and then it becomes bernoulli i think[/QUOTE]

No brother still it becomes dx/dy = x + y^2 . While the standard form of bernouli's equation is dy/dx+p(x)y=y^n. If we work out on the above equation to try to make it bernouli, the end result will be dx/dy -x =y^2. Notice the 2nd term on Right hand side is only -x here y is missing.
The paper of CE 2014 was a pathetic one. And i pray that this time it be in its limit.
Try some other method if u could.

[QUOTE=engr sarakhan;747844]No brother still it becomes dx/dy = x + y^2 . While the standard form of bernouli's equation is dy/dx+p(x)y=y^n. If we work out on the above equation to try to make it bernouli, the end result will be dx/dy -x =y^2. Notice the 2nd term on Right hand side is only -x here y is missing.
The paper of CE 2014 was a pathetic one. And i pray that this time it be in its limit.
Try some other method if u could.[/QUOTE]

But the equation you have dx/dy -x =y^2 is a Bernaouli in the variable x rather than y. Here the meanings for x and y are interchanged and we solve for x and not for y.

the form dx/dy is linear in variable x. it isn't bernoulli but it is linear with y as independent variable. its integrating factor is exp(-y).

Yeah. It is linear and not Bernouli.

Can anyone here help me out with applied mathematics?? I need it’s notes