JEE Mains · Maths · STD 12 - 9. differential equations
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 & Solution
Correct Answer
(D) \(y\sqrt {\cot \,x} = x + c\)
Step-by-step Solution
Detailed explanation
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…
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