The general solution of the differential equa | vedclass

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Question

(A) Given differential equation is $x dy - y dx = \sqrt{x^2 + y^2} dx$. Dividing both sides by $x dx$ (assuming $x 
eq 0$),we get: $\frac{dy}{dx} = \

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Given differential equation is $x dy - y dx = \sqrt{x^2 + y^2} dx$. Dividing both sides by $x dx$ (assuming $x 
eq 0$),we get: $\frac{dy}{dx} = \frac{y}{x} + \sqrt{1 + \frac{y^2}{x^2}}$. Let $y = vx$,then $\frac{dy}{dx} = v + x \frac{dv}{dx}$. Substituting these into the equation: $v + x \frac{dv}{dx} = v + \sqrt{1 + v^2}$. Subtracting $v$ from both sides: $x \frac{dv}{dx} = \sqrt{1 + v^2}$. Separating the variables: $\frac{dv}{\sqrt{1 + v^2}} = \frac{dx}{x}$. Integrating both sides: $\int \frac{dv}{\sqrt{1 + v^2}} = \int \frac{dx}{x}$. $\log |v + \sqrt{1 + v^2}| = \log |x| + \log |c|$. $v + \sqrt{1 + v^2} = cx$. Substituting $v = \frac{y}{x}$ back: $\frac{y}{x} + \sqrt{1 + \frac{y^2}{x^2}} = cx$. $\frac{y + \sqrt{x^2 + y^2}}{x} = cx$. $y + \sqrt{x^2 + y^2} = cx^2$.

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

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