Question

For the given differential equation, find the particular solution satisfying the given condition: $\frac{dy}{dx}-3y\ cot\ x = sin\ 2x; y = 2\ when\ x = \frac{\pi}{2}$

Answer 1

Answer:

$\frac{y}{sinx}=2\:sinx+c$

Step-by-step explanation:

Concept:

The solution of

$\frac{dy}{dx}+Py=Q is\\\\y.e^{\int{P}dx}=\int{Q.e^{\int{P}dx}dx}+c$

$\frac{dy}{dx}-y\:cotx=sin2x$

comparing this with

$\frac{dy}{dx}+Py=Q$

we get

P= - cotx

Q= sin2x

Integrating factor

$=e^{\int{P}dx}\\\\=e^{\int{-cotx}\:dx}\\\\=e^{-\int{cotx}\:dx}\\\\=e^{-logsinx}\\\\=e^{log(sinx)^{-1}}\\\\=(sinx)^{-1}\\\\=\frac{1}{sinx}$

The solution is

$y.e^{\int{P}dx}=\int{Q.e^{\int{P}dx}dx}+c\\\\y.(\frac{1}{sinx})=\int{sin2x.\frac{1}{sinx}\:dx}+c\\\\\frac{y}{sinx}=\int{2.sinx.cosx.\frac{1}{sinx}\:dx}+c\\\\\frac{y}{sinx}=2\int{cosx\:dx}+c\\\\\frac{y}{sinx}=2\:sinx+c$