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JEE Mains · Maths · STD 12 - 9. differential equations

વિકલ સમીકરણ \(\sqrt{1+x^{2}+y^{2}+x^{2} y^{2}}+x y \frac{d y}{d x}=0\) નું વ્યાપક ઉકેલ શોધો 

  1. A \(\sqrt{1+y^{2}}+\sqrt{1+x^{2}}=\frac{1}{2} \log _{e}\left(\frac{\sqrt{1+x^{2}}+1}{\sqrt{1+x^{2}}-1}\right)+C\)
  2. B \(\sqrt{1+y^{2}}-\sqrt{1+x^{2}}=\frac{1}{2} \log _{e}\left(\frac{\sqrt{1+x^{2}}+1}{\sqrt{1+x^{2}}-1}\right)+C\)
  3. C \(\sqrt{1+y^{2}}+\sqrt{1+x^{2}}=\frac{1}{2} \log _{e}\left(\frac{\sqrt{1+x^{2}}-1}{\sqrt{1+x^{2}}+1}\right)+C\)
  4. D \(\sqrt{1+y^{2}}-\sqrt{1+x^{2}}=\frac{1}{2} \log _{e}\left(\frac{\sqrt{1+x^{2}}-1}{\sqrt{1+x^{2}}+1}\right)+C\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\sqrt{1+y^{2}}+\sqrt{1+x^{2}}=\frac{1}{2} \log _{e}\left(\frac{\sqrt{1+x^{2}}+1}{\sqrt{1+x^{2}}-1}\right)+C\)

Step-by-step Solution

Detailed explanation

\(\sqrt{1+x^{2}+y^{2}+x^{2} y^{2}}+x y \frac{d y}{d x}=0\) \(\Rightarrow \sqrt{(1+x)^{2}\left(1+y^{2}\right)}+x y \frac{d y}{d x}=0\) \(\Rightarrow \sqrt{1+x^{2}} \sqrt{1+y^{2}}=-x y \frac{d y}{d x}\)…
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