Question

The general solution of the differential equation, $\sin\, 2x$ $\left(\frac{dy}{dx}-\sqrt{tan\,x}\right)-y=0,$ is :

• A. $y\sqrt{tan\,x}=x+c$
• B. $y\sqrt{cot\,x}=tan\,x+c$
• C. $y\sqrt{tan\,x}=cot\,x+c$
• D. $y\sqrt{cot\,x}=x+c$

Answer 1

Answer: (D)

$sin \,2x \left(\frac{dy}{dx}-\sqrt{tan\,x}\right)-y=0,$
or, $\frac{dy}{dx}=\frac{y}{sin\,2x}+\sqrt{tan\,x}$
or, $\frac{dy}{dx}-y\,cosec\, 2x=\sqrt{tan\,x}\,...\left(1\right)$
Now, integrating factor $\left(I.F\right)=_{e}\int^{-cosec\,2x}$
or, $I.F.=e^{-\frac{1}{2}log\left|tan\,x\right|}=e^{log\left(\sqrt{tan\,x}\right)^{-1}}$
$=\frac{1}{\sqrt{tan\,x}}=\sqrt{cot\,x}$
Now, general solution of eq. $\left(1\right)$ is written as
$y\left(I.F.\right)=\int Q\left(I.F.\right)dx+c$
$\therefore y\sqrt{cot\,x}=\int\sqrt{tan\,x.} \sqrt{cot\,x\,dx}+c$
$\therefore y\sqrt{cot\,x}=\int1. dx+c$
$\therefore y\sqrt{cot\,x}=x+c$