Question

Hence, the given function is the solution to the given differential equation. \[\frac{c - x}{1 + cx}\] is a solution of the differential equation \[(1+x^2)\frac{dy}{dx}+(1+y^2)=0\].

Answer 1

We have,

\[y = \frac{c - x}{1 + cx} .........(1)\]

Differentiating both sides of (1) with respect to x, we get

\[\frac{dy}{dx} = \frac{\left( 1 + cx \right)\left( - 1 \right) - \left( c - x \right)\left( c \right)}{\left( 1 + cx \right)^2}\]
\[ \Rightarrow \frac{dy}{dx} = \frac{- 1 - cx - c^2 + cx}{\left( 1 + cx \right)^2}\]
\[ \Rightarrow \frac{dy}{dx} = - \frac{1 + c^2}{\left( 1 + cx \right)^2} .............\left( 2 \right)\]
Now,
\[\left( 1 + x^2 \right)\frac{dy}{dx} + \left( 1 + y^2 \right)\]
\[ = - \left( 1 + x^2 \right)\frac{\left( 1 + c^2 \right)}{\left( 1 + cx \right)^2} + \left\{ 1 + \frac{\left( c - x \right)^2}{\left( 1 + cx \right)^2} \right\} .........\left[\text{Using }\left( 1 \right)\text{ and  }\left( 2 \right) \right]\]
\[ = - \frac{\left( 1 + x^2 \right)\left( 1 + c^2 \right)}{\left( 1 + cx \right)^2} + \frac{\left( 1 + cx \right)^2 + \left( c - x \right)^2}{\left( 1 + cx \right)^2}\]
\[ = - \frac{\left( 1 + x^2 \right)\left( 1 + c^2 \right)}{\left( 1 + cx \right)^2} + \frac{1 + 2cx + c^2 x^2 + c^2 - 2cx + x^2}{\left( 1 + cx \right)^2}\]
\[ = - \frac{\left( 1 + x^2 \right)\left( 1 + c^2 \right)}{\left( 1 + cx \right)^2} + \frac{\left( 1 + x^2 \right) + c^2 \left( 1 + x^2 \right)}{\left( 1 + cx \right)^2}\]
\[ = - \frac{\left( 1 + x^2 \right)\left( 1 + c^2 \right)}{\left( 1 + cx \right)^2} + \frac{\left( 1 + x^2 \right)\left( 1 + c^2 \right)}{\left( 1 + cx \right)^2} = 0\]
\[ \Rightarrow \left( 1 + x^2 \right)\frac{dy}{dx} + \left( 1 + y^2 \right) = 0\]

Hence, the given function is the solution to the given differential equation.