The solution of the differential equation x^2 | vedclass

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Question

(A) The given equation is a homogeneous differential equation: $\frac{dy}{dx} = \frac{x^2 + xy + y^2}{x^2} = 1 + \frac{y}{x} + \left( \frac{y}{x} \rig

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$	an^{-1} \left( \frac{y}{x} ight) = \log x + c$

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(A) The given equation is a homogeneous differential equation: $\frac{dy}{dx} = \frac{x^2 + xy + y^2}{x^2} = 1 + \frac{y}{x} + \left( \frac{y}{x} ight)^2$. Substitute $y = vx$,which implies $\frac{dy}{dx} = v + x \frac{dv}{dx}$. Substituting these into the equation: $v + x \frac{dv}{dx} = 1 + v + v^2$. Simplifying,we get $x \frac{dv}{dx} = 1 + v^2$. Separating the variables: $\frac{dv}{1 + v^2} = \frac{dx}{x}$. Integrating both sides: $\int \frac{dv}{1 + v^2} = \int \frac{dx}{x}$. This yields $	an^{-1}(v) = \log |x| + c$. Substituting $v = \frac{y}{x}$ back,we get $	an^{-1} \left( \frac{y}{x} ight) = \log |x| + c$.

The solution of the differential equation $x^2 \frac{dy}{dx} = x^2 + xy + y^2$ is

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