हल कीजिए: x dy - y dx - x (x^2 - y^2)^(1/2) dx = 0

Steps to Solve

1. Rearrange the equation Begin by isolating the terms with $dy$ and $dx$: $$ x , dy - y , dx = x \sqrt{x^2 - y^2} , dx $$

2. Divide by $x^2$ Divide both sides of the equation by $x^2$ to get it into a more manageable form: $$ \frac{x , dy - y , dx}{x^2} = \frac{x \sqrt{x^2 - y^2}}{x^2} , dx $$ $$ \frac{x , dy - y , dx}{x^2} = \frac{\sqrt{x^2 - y^2}}{x} , dx $$

3. Recognize the left side as a derivative Notice that the left side is the derivative of $\frac{y}{x}$: $$ d\left(\frac{y}{x}\right) = \frac{\sqrt{x^2 - y^2}}{x} , dx $$

4. Simplify the square root term We can simplify the square root term by factoring out $x^2$: $$ d\left(\frac{y}{x}\right) = \frac{\sqrt{x^2(1 - \frac{y^2}{x^2})}}{x} , dx $$ $$ d\left(\frac{y}{x}\right) = \frac{x\sqrt{1 - \frac{y^2}{x^2}}}{x} , dx $$ $$ d\left(\frac{y}{x}\right) = \sqrt{1 - \frac{y^2}{x^2}} , dx $$

5. Integrate by substitution Let $u = \frac{y}{x}$. Then $du = \sqrt{1 - u^2} , \frac{dx}{x}$. Rearrange to isolate $dx$: Start with $$ du = \sqrt{1-u^2} , dx $$ $$ \frac{du}{\sqrt{1 - u^2}} = \frac{dx}{x} $$ Integrate both sides $$ \int \frac{du}{\sqrt{1 - u^2}} = \int \frac{dx}{x} $$ $$ \sin^{-1}(u) = \ln|x| + C $$

6. Substitute back for $u$ Substitute $u = \frac{y}{x}$ back into the equation: $$ \sin^{-1}\left(\frac{y}{x}\right) = \ln|x| + C $$

7. Isolate $y$ Solve for $y$: $$ \frac{y}{x} = \sin(\ln|x| + C) $$ $$ y = x \sin(\ln|x| + C) $$