Question

$\sin x\frac{dy}{dx}+3y=\cos x$ , solving it gives

$(\frac{1}{3}+y){\tan }^k\frac{x}{2}=c+2\tan \frac{x}{2}-x$

What is the value of k?

• A. 1
• B. 2
• C. 5
• D. 3

Answer 1

The correct option is D 3

$\frac{dy}{dx}+3cosecx.y=tan\left(x\right)$ ...(after dividing by $sin\left(x\right)$).
Then the above differential equation is of the form,
$\frac{dy}{dx}+P\left(x\right).y=Q\left(x\right)$.
Hence
$IF=e^{\int P\left(x\right).dx}$
$=e^{\int 3\left(cosec\right(x\left)\right).dx}$
$=e^{-3ln|cosec\left(x\right)+cot\left(x\right)|}$
$=\frac{1}{\left(cosec\right(x)+cot(x){)}^3}$
$=\frac{si{n}^3\left(x\right)}{(1+cos(x){)}^3}$
$=\frac{8si{n}^3\left(\frac{x}{2}\right).co{s}^3\left(\frac{x}{2}\right)}{8co{s}^6\left(\frac{x}{2}\right)}$
$=ta{n}^3\left(\frac{x}{2}\right)$.
Now
$IF.y=\int Q\left(x\right).IF.dx$
Hence
$k=3$.