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

Question

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

dy\dx = y tan x - 2 sin x

Answer 1

Given Differential Equation :

\(\frac{dy}{dx}\) =y tan x - 2 sin x

Formula :

i) \(\int\) tan x dx = log |sec x|

ii) alog b = log b^a

iii) al^{og_a b} = b

iv) 2 sin x. cos x = sin 2x

v) \(\int\) sin x dx =  sin x

vi) 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 tan x - 2 sin x

\(\frac{dy}{dx}\) - y tan x = - 2 sin x .....eq(1)

Equation (1) is of the form

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

where' P = - tan x and Q  = - 2 sin x

Therefore, integrating factor is

General solution is

Multiplying above equation by 2,

∴ 2y. (cos x) = cos 2x + 2c

∴ 2y. (cos x) = cos 2x + c where, c = 2c

Therefore, general solution is

2y. (cos x) = cos 2x + c