The general solution of the differential equa | vedclass

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Question

(A) Given the differential equation: $(y^2 - x^2) dx = xy dy$. This can be rewritten as: $\frac{dy}{dx} = \frac{y^2 - x^2}{xy}$. This is a homogeneous

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$2x^2 \log |x| + y^2 + 2cx^2 = 0$,where $c$ is the constant of integration

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(A) Given the differential equation: $(y^2 - x^2) dx = xy dy$. This can be rewritten as: $\frac{dy}{dx} = \frac{y^2 - x^2}{xy}$. This is a homogeneous differential equation. Let $y = vx$,then $\frac{dy}{dx} = v + x \frac{dv}{dx}$. Substituting these into the equation: $v + x \frac{dv}{dx} = \frac{(vx)^2 - x^2}{x(vx)} = \frac{v^2 - 1}{v}$. $x \frac{dv}{dx} = \frac{v^2 - 1}{v} - v = \frac{v^2 - 1 - v^2}{v} = -\frac{1}{v}$. Separating the variables: $v dv = -\frac{1}{x} dx$. Integrating both sides: $\int v dv = -\int \frac{1}{x} dx$. $\frac{v^2}{2} = -\log |x| + c$. Since $v = \frac{y}{x}$,we have $\frac{y^2}{2x^2} = -\log |x| + c$. Multiplying by $2x^2$: $y^2 = -2x^2 \log |x| + 2cx^2$. Rearranging gives: $2x^2 \log |x| + y^2 - 2cx^2 = 0$. Replacing $-c$ with a new constant $C$,we get $2x^2 \log |x| + y^2 + 2Cx^2 = 0$.

The general solution of the differential equation $(y^2-x^2) dx = xy dy$ $(x \neq 0)$ is

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