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STD 12 - 9. differential equations — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 9. differential equations1 Mark
MCQ
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
    $y\sqrt {\tan \,x}  = \cot x + c$
  • ✓
    $y\sqrt {\cot \,x}  = x + c$
✓

Answer

Correct option: D.
$y\sqrt {\cot \,x}  = x + c$
d
Given, $\sin 2 x\left(\frac{d y}{d x}-\sqrt{\tan x}\right)-y=0$

or, $\frac{d y}{d x}=\frac{y}{\sin 2 x}+\sqrt{\tan x}$

or, $\frac{{dy}}{{dx}} - y\cos ec2x = \sqrt {\tan x} $          ....$(1)$

Now, integrating factor (I.F) $ = {e^{\int  -  \cos ec2x}}$

or, I.F $=e^{-\frac{1}{2} \log \tan x |}=e^{\log (\sqrt{\tan x})^{-1}}$

$=\frac{1}{\sqrt{\tan x}}=\sqrt{\cot x}$

Now, general solution of eq. (1) is written as

y$\left( {I.F.} \right)$ $=\int \mathrm{Q}(\mathrm{LF} .) d x+c$

$y \sqrt{\cot x}=\int \sqrt{\tan x} \cdot \sqrt{\cot x} d x+c$

$\therefore y \sqrt{\cot x}=\int 1 . d x+c$

$\boxed{\therefore y\sqrt {\cot x}  = x + c}$

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