Solve the following Differential Equation solvable for $x$

Question

We have the following Differential Equation: $x+\frac{p}{\sqrt{1+p^2}}=a$ where $p$ is the first derivative of $y$ with respect to $x$.

I am not able to start the problem. I know that this is solvable for $x$ but after that it is getting difficult to derivative. Please help.

Answer 1

Hint

Consider the equation to be $$x+\frac{y'}{\sqrt{1+y'^2}}=a$$ Change variable $t=x-a$. Square and solve for $y'^2$ to get $$y'=\pm \frac{t}{\sqrt{1-t^2}}$$ which does not seem to be too difficult.

Answer 2

$$x+\frac{p}{\sqrt{1+p^2}}=a \implies \\ p^2=\frac {(x-a)^2}{1-(x-a)^2} \implies \\ y'=\pm \frac {u}{\sqrt {1-u^2}}$$

Where $u=x-a$

Upon integration you find your $y$ as function of $u=x-a$