Question

Show that the general solution of the differential equation:

`dy/dx` = y cot 2x is log y = `1/2 log|sin2x| + C`

Answer 1

Given, `dy/dx` = y cot 2x

We separate the variables by writing:

`1/y dy` = cot 2x dx

Integrate both sides:

`int 1/y dy = int cot 2x dx`

Integrating the left side,

`int 1/y dy` = log |y|

Integrating the right side,

cot 2x = `(cos 2x)/(sin 2x)`

Let u = sin⁡ 2x

⇒ du = 2 cos ⁡2x dx

⇒ cos ⁡2x dx = `1/2` du

So, `int cot 2x dx = int(cos 2x)/(sin 2x)dx`

= `int 1/u * 1/2 du`

= `1/2 int 1/u du`

= `1/2 log |u|`

= `1/2 log |sin 2x|`

Thus, `log |y| = 1/2 log |sin 2x| + C`