The equation of the curve satisfying x dy - y | vedclass

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(D) Given the differential equation: $x dy - y dx = \sqrt{x^2 - y^2} dx$. Divide by $dx$: $x \frac{dy}{dx} - y = \sqrt{x^2 - y^2}$. Substitute $y = vx

Vedclass · Accepted Answer

$y = x \sin(\ln x)$

Vedclass · Answer

(D) Given the differential equation: $x dy - y dx = \sqrt{x^2 - y^2} dx$. Divide by $dx$: $x \frac{dy}{dx} - y = \sqrt{x^2 - y^2}$. Substitute $y = vx$,so $\frac{dy}{dx} = v + x \frac{dv}{dx}$. $x(v + x \frac{dv}{dx}) - vx = \sqrt{x^2 - v^2x^2}$. $vx + x^2 \frac{dv}{dx} - vx = x \sqrt{1 - v^2}$. $x^2 \frac{dv}{dx} = x \sqrt{1 - v^2} \implies \frac{dv}{\sqrt{1 - v^2}} = \frac{dx}{x}$. Integrating both sides: $\int \frac{dv}{\sqrt{1 - v^2}} = \int \frac{dx}{x} \implies \sin^{-1}(v) = \ln|x| + C$. Substitute $v = \frac{y}{x}$: $\sin^{-1}(\frac{y}{x}) = \ln|x| + C$. Given $y(1) = 0$,at $x = 1, y = 0$: $\sin^{-1}(0) = \ln(1) + C \implies 0 = 0 + C \implies C = 0$. Thus,$\sin^{-1}(\frac{y}{x}) = \ln x \implies \frac{y}{x} = \sin(\ln x) \implies y = x \sin(\ln x)$.

The equation of the curve satisfying $x dy - y dx = \sqrt{x^2 - y^2} dx$ with the condition $y(1) = 0$ is:

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