If \(\mathrm{U}(x) = x+\sqrt{1+x^{2}}\), then solution of the differential equation \(\displaystyle\frac{dy}{dx}+\sqrt{\frac{1+y^{2}}{1+x^{2}}}=0\), is :
(where c is arbitrary constant)
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\mathrm{U}(x)\mathrm{U}(y) = \mathrm{U}(c)\)
The variables separate:
\( \displaystyle \frac{dy}{\sqrt{1+y^{2}}} + \frac{dx}{\sqrt{1+x^{2}}} = 0 \)
Using \(\displaystyle\int\frac{dx}{\sqrt{1+x^{2}}} = \log\left|x+\sqrt{1+x^{2}}\right|\), and writing the constant as log k:
\( \log\left(y+\sqrt{1+y^{2}}\right) + \log\left(x+\sqrt{1+x^{2}}\right) = \log k \)
(both brackets are positive for every real value, so no modulus is needed). Hence
\( \left(x+\sqrt{1+x^{2}}\right)\left(y+\sqrt{1+y^{2}}\right) = k \)
\(\mathrm{U}(x)\mathrm{U}(y) = k\)
U takes every positive value, so the positive constant k can be written as U(c) for some constant c:
\(\mathrm{U}(x)\mathrm{U}(y) = \mathrm{U}(c)\)