Question

Show that y = $\frac{c-x}{1+cx}$ is a solution of the differential equation (1 + x²) $\frac{dy}{dx}$ + (1 + y²) = 0.

Answer 1

We have,

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

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

$$\begin{gathered} \frac{dy}{dx}=\frac{\left(1+cx\right)\left(-1\right)-\left(c-x\right)\left(c\right)}{{\left(1+cx\right)}^2} \\ \Rightarrow \frac{dy}{dx}=\frac{-1-cx-c^2+cx}{{\left(1+cx\right)}^2} \\ \Rightarrow \frac{dy}{dx}=-\frac{1+c^2}{{\left(1+cx\right)}^2}...\left(2\right) \end{gathered}$$

Now,

$$\begin{gathered} \left(1+x^2\right)\frac{dy}{dx}+\left(1+y^2\right) \\ =-\left(1+x^2\right)\frac{\left(1+c^2\right)}{{\left(1+cx\right)}^2}+\left\{1+\frac{{\left(c-x\right)}^2}{{\left(1+cx\right)}^2}\right\}\left[\mathrm{Using}\left(1\right)\mathrm{and}\left(2\right)\right] \\ =-\frac{\left(1+x^2\right)\left(1+c^2\right)}{{\left(1+cx\right)}^2}+\frac{{\left(1+cx\right)}^2+{\left(c-x\right)}^2}{{\left(1+cx\right)}^2} \\ =-\frac{\left(1+x^2\right)\left(1+c^2\right)}{{\left(1+cx\right)}^2}+\frac{1+2cx+c^2{x}^2+c^2-2cx+x^2}{{\left(1+cx\right)}^2} \\ =-\frac{\left(1+x^2\right)\left(1+c^2\right)}{{\left(1+cx\right)}^2}+\frac{\left(1+x^2\right)+c^2\left(1+x^2\right)}{{\left(1+cx\right)}^2} \\ =-\frac{\left(1+x^2\right)\left(1+c^2\right)}{{\left(1+cx\right)}^2}+\frac{\left(1+x^2\right)\left(1+c^2\right)}{{\left(1+cx\right)}^2}=0 \\ \Rightarrow \left(1+x^2\right)\frac{dy}{dx}+\left(1+y^2\right)=0 \end{gathered}$$

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