Given Differential Equation :
\(\frac{dy}{dx}\) + 2y tan x = sin x
Formula :
i) \(\int\) tan x dx = log |sec x|
ii) alog b = log ba
iii) aloga b = b
iv) \(\int\) \((\frac{-1}{x^2})\) dx = \(\frac{1}{x}\)
v) 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
\(\frac{dy}{dx}\) + 2y tan x = sin x ....eq(1)
Equation (1) is of the form
\(\frac{dy}{dx} \, + Py \, =Q\)
where, P = 2tan x and Q = sin x
Therefore, integrating factor is
General solution is
Substituting I in eq(2),
Multiplying above equation by cos2x,
Therefore, general solution is
y = cos x + c(cos2 x)