The solution of the differential equation 3 x | vedclass

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(A) Given differential equation: $3 x y' - 3 y + \sqrt{x^2 - y^2} = 0$. Dividing by $x$: $3 y' - 3 \frac{y}{x} + \sqrt{1 - (\frac{y}{x})^2} = 0$. Let

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$3 \cos^{-1}\left(\frac{y}{x}\right) = \ln |x|$

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(A) Given differential equation: $3 x y' - 3 y + \sqrt{x^2 - y^2} = 0$. Dividing by $x$: $3 y' - 3 \frac{y}{x} + \sqrt{1 - (\frac{y}{x})^2} = 0$. Let $y = vx$, then $y' = v + x \frac{dv}{dx}$. Substituting: $3(v + x \frac{dv}{dx}) - 3v + \sqrt{1 - v^2} = 0$. $3v + 3x \frac{dv}{dx} - 3v + \sqrt{1 - v^2} = 0$. $3x \frac{dv}{dx} = -\sqrt{1 - v^2}$. Separating variables: $\frac{-dv}{\sqrt{1 - v^2}} = \frac{1}{3} \frac{dx}{x}$. Integrating both sides: $\int \frac{-dv}{\sqrt{1 - v^2}} = \int \frac{1}{3} \frac{dx}{x}$. $\cos^{-1}(v) = \frac{1}{3} \ln |x| + C$. Since $y(1) = 1$, $v = \frac{y}{x} = 1$ when $x = 1$. $\cos^{-1}(1) = \frac{1}{3} \ln(1) + C \Rightarrow 0 = 0 + C \Rightarrow C = 0$. Thus, $\cos^{-1}(\frac{y}{x}) = \frac{1}{3} \ln |x|$, which implies $3 \cos^{-1}(\frac{y}{x}) = \ln |x|$.

The solution of the differential equation $3 x y' - 3 y + (x^2 - y^2)^{1/2} = 0$, satisfying the condition $y(1) = 1$ is

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