JEE Mains · Maths · STD 12 - 9. differential equations
If a curve \(\mathrm{y}=\mathrm{f}(\mathrm{x}),\) passing through the point \((1,2),\) is the solution of the differential equation, \(2 \mathrm{x}^{2} \mathrm{dy}=\left(2 \mathrm{xy}+\mathrm{y}^{2}\right) \mathrm{dx},\) then \(\mathrm{f}\left(\frac{1}{2}\right)\) is equal to
- A \(\frac{1}{1-\log _{e} 2}\)
- B \(\frac{1}{1+\log _{e} 2}\)
- C \(\frac{-1}{1+\log _{e} 2}\)
- D \(1+\log _{e} 2\)
Answer & Solution
Correct Answer
(B) \(\frac{1}{1+\log _{e} 2}\)
Step-by-step Solution
Detailed explanation
\(2 x^{2} d y=\left(2 x y+y^{2}\right) d x\) \(\Rightarrow \frac{d y}{d x}=\frac{2 x y+y^{2}}{2 x^{2}}\{\) Homogeneous D.E. \(\}\) \(\left[\begin{array}{l}\text { let } y=x t \\ \Rightarrow \frac{d y}{d x}=t+x \frac{d t}{d x}\end{array}\right\}\)…
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