The solution of x dy - y dx = sqrt{x^2 + y^2} | vedclass

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Question

(A) Given the differential equation $x dy - y dx = \sqrt{x^2 + y^2} dx$.Rearranging the terms,we get $\frac{dy}{dx} = \frac{y + \sqrt{x^2 + y^2}}{x}$.

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

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(A) Given the differential equation $x dy - y dx = \sqrt{x^2 + y^2} dx$.Rearranging the terms,we get $\frac{dy}{dx} = \frac{y + \sqrt{x^2 + y^2}}{x}$.This is a homogeneous differential equation. Let $y = vx$,then $\frac{dy}{dx} = v + x \frac{dv}{dx}$.Substituting these into the equation: $v + x \frac{dv}{dx} = \frac{vx + \sqrt{x^2 + v^2x^2}}{x} = v + \sqrt{1 + v^2}$.Simplifying,we have $x \frac{dv}{dx} = \sqrt{1 + v^2}$.Separating the variables: $\int \frac{dv}{\sqrt{1 + v^2}} = \int \frac{dx}{x}$.Integrating both sides: $\ln(v + \sqrt{v^2 + 1}) = \ln|x| + C = \ln|cx|$.Thus,$v + \sqrt{v^2 + 1} = cx$.Substituting $v = \frac{y}{x}$: $\frac{y}{x} + \sqrt{\frac{y^2}{x^2} + 1} = cx \Rightarrow y + \sqrt{x^2 + y^2} = cx^2$.Given $y(\sqrt{3}) = 1$,we have $1 + \sqrt{3 + 1} = c(\sqrt{3})^2 \Rightarrow 1 + 2 = 3c \Rightarrow c = 1$.Therefore,the solution is $y + \sqrt{x^2 + y^2} = x^2$.

The solution of $x dy - y dx = \sqrt{x^2 + y^2} dx$ when $y(\sqrt{3}) = 1$ is

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