The solution of the differential equation 1+x^ 2 +y^ 2 +x^ 2 y^ 2 +x y d y d x =0 is,

Mathematics · JEE Main · NTA Exams — Differential Equations

1. \(\sqrt{1+x^{2}}-\frac{1}{2} \log \left|\frac{\sqrt{1+x^{2}}-1}{\sqrt{1+x^{2}}+1}\right|+\sqrt{1+y^{2}}=\text { constant }\)
2. \(\sqrt{1+x^{2}}-\frac{1}{2} \log \left|\frac{\sqrt{1+x^{2}}+1}{\sqrt{1+x^{2}}-1}\right|-\sqrt{1+y^{2}}=\text { constant }\)
3. \(\sqrt{1+x^{2}}+\frac{1}{2} \log \left|\frac{\sqrt{1+x^{2}}+1}{\sqrt{1+x^{2}}-1}\right|-\sqrt{1+y^{2}}=\text { constant }\)
4. \(\sqrt{1+x^{2}}+\frac{1}{2} \log \left|\frac{\sqrt{1+x^{2}}-1}{\sqrt{1+x^{2}}+1}\right|+\sqrt{1+y^{2}}=\text { constant }\)

Answer

(A) 1+x^ 2 - 1 2 | 1+x^ 2 -1 1+x^ 2 +1 |+ 1+y^ 2 = constant

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