The solution of (x^2+y^2) dx = 2xy dy is: | vedclass

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(A) Given differential equation is $(x^2+y^2) dx = 2xy dy$. Rearranging,we get $\frac{dy}{dx} = \frac{x^2+y^2}{2xy}$. This is a homogeneous differenti

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$c(x^2-y^2) = x$

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(A) Given differential equation is $(x^2+y^2) dx = 2xy dy$. Rearranging,we get $\frac{dy}{dx} = \frac{x^2+y^2}{2xy}$. This is a homogeneous differential equation. Let $y = vx$,then $\frac{dy}{dx} = v + x \frac{dv}{dx}$. Substituting these into the equation: $v + x \frac{dv}{dx} = \frac{x^2 + (vx)^2}{2x(vx)} = \frac{x^2(1+v^2)}{2x^2v} = \frac{1+v^2}{2v}$. $x \frac{dv}{dx} = \frac{1+v^2}{2v} - v = \frac{1+v^2-2v^2}{2v} = \frac{1-v^2}{2v}$. Separating variables: $\frac{2v}{1-v^2} dv = \frac{1}{x} dx$. Integrating both sides: $\int \frac{2v}{1-v^2} dv = \int \frac{1}{x} dx$. Let $1-v^2 = t$,then $-2v dv = dt$. $-\int \frac{dt}{t} = \log|x| + \log|c|$. $-\log|1-v^2| = \log|cx|$. $\log|\frac{1}{1-v^2}| = \log|cx| \Rightarrow \frac{1}{1-v^2} = cx$. Substituting $v = \frac{y}{x}$: $\frac{1}{1-(y^2/x^2)} = cx \Rightarrow \frac{x^2}{x^2-y^2} = cx$. $x = c(x^2-y^2)$.

The solution of $(x^2+y^2) dx = 2xy dy$ is:

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