Question

Prove that the following D.E is exact equation, find the solution:

dy = (y/x - cosec^2 (y/x)) dx

I know it's homogeneous but that's the question.

Answer 1

what does he mean by exact equation?

Answer 2

\[(\delta f / \delta x) * dx + (\delta f / \delta y) * dy = 0\]

Answer 3

for an exact equation

Answer 4

an example of an exact equation would be

xy' + y = e^x

Answer 5

this can be rearranged to

x dy + (y - e^x) dx = 0

we see that the partial derivatives are both = 1 so the equation is exact.

Answer 6

If we let u= y/x
and thus u x = y from which we get dy= u dx + x du
your equation can be written as
\[ u \ dx + x \ du = u - \csc^2 u \ dx \\
x \ du = -\csc^2 u \ dx \]
or
\[ x\ du = - \frac{1}{\sin^2 u}\ du \\
\frac{1}{x} \ dx +\sin^2 u \ du =0\]
the equation is exact if
\[ \frac{\partial}{\partial u} \frac{1}{x} =\frac{\partial}{\partial x} \sin^2 u\]
which it is, 0=0

Answer 7

\[\mathrm dy = \left(\frac{y}{x}-\csc^2\frac{y}{x}\right)\,\mathrm dx\iff \mathrm dy+\left(\csc^2\frac{y}{x}-\frac{y}{x}\right)\,\mathrm dx=0\]The ODE is exact if
\[\frac{\partial}{\partial x}1=\frac{\partial}{\partial y}\left[\csc^2\frac{y}{x}-\frac{y}{x}\right]\]The LHS is clearly \(0\), so for the ODE to be exact this must also be the case for the RHS. You have
\[\frac{\partial}{\partial y}\left[\csc^2\frac{y}{x}-\frac{y}{x}\right]=-\frac{1}{x}\left(1+2\csc^2\frac{y}{x}\cot\frac{y}{x}\right)\neq0\]so the equation, as is, is *not* exact, but can be made to be with a substitution, as @phi has shown.

Answer 8

@phi, the best.

@HolsterEmission, Thanks too bro ^^