Solve the differential equation given below:f | vedclass

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(A) Given differential equation:$\frac{x dy}{dx} = y + \sqrt{x^2 + y^2}$$\Rightarrow \frac{dy}{dx} = \frac{y + \sqrt{x^2 + y^2}}{x}$This is a homogene

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

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(A) Given differential equation:$\frac{x dy}{dx} = y + \sqrt{x^2 + y^2}$$\Rightarrow \frac{dy}{dx} = \frac{y + \sqrt{x^2 + y^2}}{x}$This is a homogeneous differential equation. Let $y = vx$,then $\frac{dy}{dx} = v + x \frac{dv}{dx}$.Substituting these into the equation:$v + x \frac{dv}{dx} = \frac{vx + \sqrt{x^2 + v^2x^2}}{x} = v + \sqrt{1 + v^2}$$x \frac{dv}{dx} = \sqrt{1 + v^2}$Separating the variables:$\int \frac{dv}{\sqrt{1 + v^2}} = \int \frac{dx}{x}$Integrating both sides:$\ln|v + \sqrt{1 + v^2}| = \ln|x| + \ln|c|$$\ln|v + \sqrt{1 + v^2}| = \ln|cx|$$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$Or $x^2 = \frac{1}{c} [y + \sqrt{x^2 + y^2}]$,which can be written as $x^2 = C[y + \sqrt{x^2 + y^2}]$.Thus,option $A$ is correct.

Solve the differential equation given below:
$\frac{x dy}{dx} = y + \sqrt{x^2 + y^2}$

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Solve the differential equation given below:$\frac{x dy}{dx} = y + \sqrt{x^2 + y^2}$