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

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

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