Find a particular solution satisfying the given condition for differential equations. ​ for + y cot x = 4x cosec x, given that y = 0 when x = π /2.

Question

Find a particular solution satisfying the given condition for \(\frac{dy}{dx}\) + y cot x = 4x cosec x, given that y = 0 when x \(\frac{\pi}{2}\)differential equations.

dy\dx+ y cot x = 4x cosec x

Answer 1

Given Differential Equation :

\(\frac{dy}{dx}\) + y cot x = 4x cosec x

Formula :

i) \(\int\) cot x dx = log|sin x|

ii) a^{log_a b} = b

iii) \(\int\) xⁿ dx = \(\frac{x^{n+1}}{n+1}\)

iv) 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 = 4x cosec x …eq(1)

Equation (1) is of the form

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

where, P = cot x and Q = 4x cosec x

Therefore, integrating factor is

General solution is

Therefore general equation is

y. (sin x) = 2x² + c

For particular solution put y=0 and x = \(\frac{\pi}{2}\) in above equation,

Therefore, particular solution is