Find the general solution for differential equations. dy\dx y cot x = sin 2x

Question

Find the general solution for \(\frac{dy}{dx}\) = y cot x = sin 2x differential equations.

dy\dx y cot x = sin 2x

Answer 1

Given Differential Equation :

\(\frac{dy}{dx}\) = y cot x = sin 2x

Formula :

i) \(\int\) cot x dx = log |sin x|

ii) a^{log_a b} = b

iii) \(\int\) u. v dx = u. \(\int\) v dx - \(\int\) \((\frac{du}{dx}. \int v\,dx)\) dx

iv) \(\int\) sin x dx = - cos x

v) \(\frac{d}{dx}\) (sin x) = cos x

vi) 2 sin x. cos x = sin 2x

vii) cos 2x = (cos² x - sin² x) 

viii) General solution :

For the differential equation in the form of

\(\frac{dy}{dx} \, +\, Py = Q\)

General solution is given by,

y. (I.F.) = \(\int\) Q.(I.F.) dx + c

Where, integrating factor,

I. F. = \(e^\int \,p\,dx\)

Given differential equation is

\(\frac{dy}{dx}\) + y cot x = sin 2x …eq(1)

Equation (1) is of the form

\(\frac{dy}{dx} \, +\, Py = Q\)

where, P = cot x and Q = sin 2x

Therefore, integrating factor is

General solution is

Again let, u=cos 2x & v=cos x

Substituting I in eq(2),

∴ y. (sin x) = \(\frac{2}{3}\) sin³ x + c

Therefore, general solution is

y. (sin x) = \(\frac{2}{3}\) sin³ x + c