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

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(A) Given the differential equation: $(x^2+y^2) dx = 2xy dy$Rearranging,we get: $\frac{dy}{dx} = \frac{x^2+y^2}{2xy}$Since this is a homogeneous diffe

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

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(A) Given the differential equation: $(x^2+y^2) dx = 2xy dy$Rearranging,we get: $\frac{dy}{dx} = \frac{x^2+y^2}{2xy}$Since this is a homogeneous differential equation,substitute $y = vx$,which implies $\frac{dy}{dx} = v + x \frac{dv}{dx}$.Substituting these into the equation: $v + x \frac{dv}{dx} = \frac{x^2 + (vx)^2}{2x(vx)} = \frac{x^2(1+v^2)}{2x^2v} = \frac{1+v^2}{2v}$Subtracting $v$ from both sides: $x \frac{dv}{dx} = \frac{1+v^2}{2v} - v = \frac{1+v^2-2v^2}{2v} = \frac{1-v^2}{2v}$Separating the 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 $u = 1-v^2$,then $du = -2v dv$. The integral becomes: $-\int \frac{1}{u} du = \ln|x| + C$$-\ln|1-v^2| = \ln|x| + \ln|c|$$-\ln|1 - (y/x)^2| = \ln|cx|$$-\ln|\frac{x^2-y^2}{x^2}| = \ln|cx|$$\ln|\frac{x^2}{x^2-y^2}| = \ln|cx|$$\frac{x^2}{x^2-y^2} = cx$$x = c(x^2-y^2)$

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

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