Question

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

• A. $y\sqrt{tanx}=x+c$
• B. $y\sqrt{cotx}=tanx+c$
• C. $y\sqrt{tanx}=cotx+c$
• D. $y\sqrt{cotx}=x+c$

Answer 1

Answer: (D)

$sin2x\left(\frac{dy}{dx}-\sqrt{tanx}\right)-y=0,$
or, $\frac{dy}{dx}=\frac{y}{sin2x}+\sqrt{tanx}$
or, $\frac{dy}{dx}-ycosec2x=\sqrt{tanx}...\left(1\right)$
Now, integrating factor $\left(I.F\right){=}_e{\int }^{-cosec2x}$
or, $I.F.=e^{-\frac{1}{2}log|tanx|}=e^{log{\left(\sqrt{tanx}\right)}^{-1}}$
$=\frac{1}{\sqrt{tanx}}=\sqrt{cotx}$
Now, general solution of eq. $\left(1\right)$ is written as
$y\left(I.F.\right)=\int Q\left(I.F.\right)dx+c$
$\therefore y\sqrt{cotx}=\int \sqrt{tanx.}\sqrt{cotxdx}+c$
$\therefore y\sqrt{cotx}=\int 1.dx+c$
$\therefore y\sqrt{cotx}=x+c$