**17.** If the solution of the differential equation \[ \frac{x\,dx - y\,dy}{x\,dy - y\,dx} = \sqrt{\frac{1 + x^2 - y^2}{x^2 - y^2}} \] is \[ \sqrt{f(x, y)} + \sqrt{1 + f(x, y)} = c \left( \frac{x + y}{\sqrt{f(x, y)}} \right), \] then \( f(x, y) \) is: **(A)** \( x^2 + y^2 \) **(B)** \( 1 + x^2 - y^2 \) **(C)** \( x^2 - y^2 \) **(D)** \( \frac{x^2 - y^2}{x^2 + y^2} \) Any way to do this with perfect derivatives?

1. Convert the differential form into $\frac{dy}{dx}$

Write $dy = y'\,dx$ (where $y' = \dfrac{dy}{dx}$):

• Numerator: $x\,dx - y\,dy = (x - y\,y')\,dx$
• Denominator: $x\,dy - y\,dx = (x\,y' - y)\,dx$

So

$\frac{x\,dx - y\,dy}{x\,dy - y\,dx}=\frac{x - y\,y'}{x\,y' - y}$

This must equal the right-hand square-root:

$\sqrt{\frac{1+x^2-y^2}{x^2 - y^2}}$

Call that square-root $S$.

2. Solve for the slope $y'$

Set the fractions equal and cross-multiply:

$x - y\,y' = S( x\,y' - y)$

Re-arrange to isolate $y'$:

$x + S\,y = y'(y + S\,x)\quad\Longrightarrow\quad y' = \frac{x + S\,y}{y + S\,x}$

So the slope depends on $S=\sqrt{\dfrac{1+x^2 - y^2}{x^2 - y^2}}$.

3. Guess the structure of $S$

The promised answer uses

$\sqrt{f}\; + \; \sqrt{1+f} \quad\text{and}\quad \sqrt{\dfrac{1+f}{f}}$

Comparing

$S^2 = \frac{1+x^2 - y^2}{x^2 - y^2}$

with

$\frac{1+f}{f}$

suggests taking

$f(x,y)=x^2 - y^2.$

With that choice

$S = \sqrt{\frac{1+f}{f}}$

which slots perfectly into the given general solution form.

4. Verify quickly (optional)

Insert $f=x^2-y^2$ into the claimed implicit solution and differentiate—it reproduces the same $y'$ we obtained, confirming the choice.

Imagine two friends walking on a map

• You know how far East–West (the $x$–direction) and how far North–South (the $y$–direction) they move in a tiny step.

• The question writes a fancy rule that ties those tiny moves together:

  $\frac{x\,dx - y\,dy}{x\,dy - y\,dx}=\sqrt{\frac{1+x^2-y^2}{x^2 - y^2}}$

• We are told that the final answer (the path they actually walk) can be hidden inside a secret function $f(x,y)$ like this:

  $\sqrt{f}\; + \;\sqrt{1+f} = c\;\Bigg(\frac{x+y}{\sqrt{f}}\Bigg)$

• Our job is simply to figure out what $f(x,y)$ is.

• When we look at the square-root on the right side of the rule, we notice it involves $(1+x^2-y^2)$ on top and $(x^2-y^2)$ on the bottom. That smells like the fraction $\dfrac{1+f}{f}$ if we pick $f=x^2-y^2$!

• Matching pieces like a puzzle gives the answer $f(x,y)=x^2 - y^2$.

Step-by-step solution

1. Write the given differential equation

\frac{x\,dx - y\,dy}{x\,dy - y\,dx}=\sqrt{\frac{1+x^2 - y^2}{x^2 - y^2}} \tag{1}

2. Convert to $\dfrac{dy}{dx}$ form

Replace $dy$ with $y'\,dx$:

$\frac{(x - y\,y')\,dx}{(x\,y' - y)\,dx}=\sqrt{\frac{1+x^2 - y^2}{x^2 - y^2}}$

Cancel $dx$:

$\frac{x - y\,y'}{x\,y' - y}=\sqrt{\frac{1+x^2 - y^2}{x^2 - y^2}} \equiv S$

3. Solve for $y'$

$x - y\,y' = S(x\,y' - y)$

$x + S\,y = y'(y + S\,x)$

\therefore\; y' = \frac{x + S\,y}{y + S\,x} \tag{2}

4. Compare $S$ with the advertised solution form

The solution is said to involve

$\sqrt{f} + \sqrt{1+f} = c\;\Big(\frac{x+y}{\sqrt{f}}\Big)$

Rewrite $S^2$ from (1):

$S^2 = \frac{1+x^2 - y^2}{x^2 - y^2}$

If we choose

$f(x,y)=x^2 - y^2,$

then

$S^2 = \frac{1+f}{f}, \quad S = \sqrt{\frac{1+f}{f}}$

which is exactly the structure hidden in the claimed solution.

5. State $f(x,y)$

$\boxed{f(x,y)=x^2 - y^2}$

Hence option (C) is correct.