The solution of the differential equation x d | vedclass

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(B) Given the differential equation: $x dy - y dx = \sqrt{x^2+y^2} dx$ $\ldots$ $(i)$Dividing by $x dx$ (assuming $x 
eq 0$), we get: $x \frac{dy}{dx

Vedclass · Accepted Answer

$\left(x^2-y\right)^2=x^2+y^2$

Vedclass · Answer

(B) Given the differential equation: $x dy - y dx = \sqrt{x^2+y^2} dx$ $\ldots$ $(i)$Dividing by $x dx$ (assuming $x 
eq 0$), we get: $x \frac{dy}{dx} - y = \sqrt{x^2+y^2}$.Let $y = vx$, then $\frac{dy}{dx} = v + x \frac{dv}{dx}$.Substituting into the equation: $x(v + x \frac{dv}{dx}) - vx = \sqrt{x^2 + (vx)^2}$.$vx + x^2 \frac{dv}{dx} - vx = x \sqrt{1+v^2}$.$x^2 \frac{dv}{dx} = x \sqrt{1+v^2} \Rightarrow \frac{dv}{\sqrt{1+v^2}} = \frac{dx}{x}$.Integrating both sides: $\ln(v + \sqrt{1+v^2}) = \ln x + C$.Substituting $v = \frac{y}{x}$: $\ln(\frac{y}{x} + \sqrt{1 + \frac{y^2}{x^2}}) = \ln x + C$.$\ln(\frac{y + \sqrt{x^2+y^2}}{x}) = \ln x + C \Rightarrow \ln(\frac{y + \sqrt{x^2+y^2}}{x^2}) = C$.Given $y=1$ when $x=\sqrt{3}$: $\ln(\frac{1 + \sqrt{3+1}}{3}) = C \Rightarrow \ln(\frac{1+2}{3}) = C \Rightarrow \ln(1) = C \Rightarrow C = 0$.Thus, $\ln(\frac{y + \sqrt{x^2+y^2}}{x^2}) = 0 \Rightarrow \frac{y + \sqrt{x^2+y^2}}{x^2} = 1$.$y + \sqrt{x^2+y^2} = x^2 \Rightarrow \sqrt{x^2+y^2} = x^2 - y$.Squaring both sides: $x^2 + y^2 = (x^2 - y)^2$.

The solution of the differential equation $x dy - y dx = \sqrt{x^2+y^2} dx$, given that $y=1$ when $x=\sqrt{3}$, is

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