Question

Find a particular solution of the differential equation $\frac{dy}{dx} +y\ cot\ x = 4x\ cosec\ x(x \neq 0), given\ that\ y = 0\ when\ x=\frac{\pi}{2}$

Answer 1

Answer:

y. sin(x) = 2x² + c

Step-by-step explanation:

Hi,

Given that ${\frac{dy}{dx}$ + y cot x = 4x cosec x,

By comparing the above equation to Linear differential

equation,

${\frac{dy}{dx}}$ + y P = Q,--------(1)

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

As we know, the solution to equation (1), is given by

y.${e^{{\int Pdx}$ = ${\int Qe^{Pdx}}$ + c,-----(2)

But ${e^{{\int Pdx}$ = ${e^{{\int cot x dx}$

= ${e^{㏑ sin x}$

= sin x.

Hence, equation (2), will become

y.sinx = ${\int Qsin x}$ + c,

= ${\int 4x cosec x* sin x dx}}$ + c,

= ${\int 4x dx}}$ + c,

= 2x² + c.

Hence, the particular solution for the given equation

is y. sin(x) = 2x² + c, where c is an arbitrary constant.

Hope, it helps !