The solution of the differential equation x , | vedclass

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(B) Given 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}$$x \frac{dy}{dx}

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$y + \sqrt{x^2 + y^2} = cx^2$

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(B) Given 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}$$x \frac{dy}{dx} = y + \sqrt{x^2 + y^2}$$\frac{dy}{dx} = \frac{y + \sqrt{x^2 + y^2}}{x}$This is a homogeneous differential equation. Let $y = vx$,then $\frac{dy}{dx} = v + x \frac{dv}{dx}$.Substituting these into the equation:$v + x \frac{dv}{dx} = \frac{vx + \sqrt{x^2 + v^2x^2}}{x} = v + \sqrt{1 + v^2}$$x \frac{dv}{dx} = \sqrt{1 + v^2}$Separating variables: $\frac{dv}{\sqrt{1 + v^2}} = \frac{dx}{x}$Integrating both sides: $\ln|v + \sqrt{1 + v^2}| = \ln|x| + \ln|c|$$v + \sqrt{1 + v^2} = cx$Substitute $v = \frac{y}{x}$:$\frac{y}{x} + \sqrt{1 + \frac{y^2}{x^2}} = cx$$\frac{y + \sqrt{x^2 + y^2}}{x} = cx$$y + \sqrt{x^2 + y^2} = cx^2$

The solution of the differential equation $x \, dy - y \, dx = \sqrt{x^2 + y^2} \, dx$ is

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