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`        dy/dx +(1/x)  tan y = (1/x2)tany.siny`
10 years ago

```							dy/dx +(1/x)  tan y = (1/x2)tany.siny
divide the eqn. by tan y.sin y
so eqn comes to be
(1/tan y.sin y)*dy/dx + 1/ x.sin y = 1/x2
put 1/sin y = t
(-cosy / sin2y)*dy/dx = dt/dx
on substituting,we get
-dt/dx + t/x = 1/x2
or dt/dx - t/x = -1/x2
which is of form dy/dx +P(x).y = Q(x)  with P(x) = -1/x and  Q(x) = -1/x2
solution is t.eintegral P(x) = integral ( Q(x)*eintegral P(x))dx +C
soln is:
t .e -lnx =  -1/x2 .e -lnx dx
t/x = 1/2x2 +C
i.e., 1/x. sin y = 1/(2x2) +C
```
10 years ago
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