JEE Mains · Maths · STD 12 - 9. differential equations
Let \(C_{1}\) be the curve obtained by the solution of differential equation \(2 xy \frac{ dy }{ dx }= y ^{2}- x ^{2}, x > 0\) Let the curve \(C _{2}\) be the solution of \(\frac{2 x y}{x^{2}-y^{2}}=\frac{d y}{d x} .\) If both the curves pass through \((1,1),\) then the area enclosed by the curves \(C_{1}\) and \(C _{2}\) is equal to :
- A \(\pi-1\)
- B \(\frac{\pi}{2}-1\)
- C \(\pi + 1\)
- D \(\frac{\pi}{4}+1\)
Answer & Solution
Correct Answer
(B) \(\frac{\pi}{2}-1\)
Step-by-step Solution
Detailed explanation
\(\frac{ dy }{ dx }=\frac{ y ^{2}- x ^{2}}{2 xy }, \quad x \in(0, \infty)\) put \(y = v x\) \(x \frac{d v}{d x}+v=\frac{v^{2}-1}{2 v}\) \(\frac{2 v}{v^{2}+1} d v=-\frac{d x}{x}\) Integrate, \(\ln \left(v^{2}+1\right)=-\ln x+C\)…
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