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(B) Given differential equation is $(x^2 + y^2)dx = 2xydy$.This can be rewritten as $\frac{dy}{dx} = \frac{x^2 + y^2}{2xy}$.This is a homogeneous diff

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$x^2 - y^2 = cx$

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(B) Given differential equation is $(x^2 + y^2)dx = 2xydy$.This can be rewritten as $\frac{dy}{dx} = \frac{x^2 + y^2}{2xy}$.This is a homogeneous differential equation. Let $y = vx$,then $\frac{dy}{dx} = v + x\frac{dv}{dx}$.Substituting these into the equation: $v + x\frac{dv}{dx} = \frac{x^2 + v^2x^2}{2x(vx)} = \frac{x^2(1 + v^2)}{2x^2v} = \frac{1 + v^2}{2v}$.$x\frac{dv}{dx} = \frac{1 + v^2}{2v} - v = \frac{1 + v^2 - 2v^2}{2v} = \frac{1 - v^2}{2v}$.Separating the variables: $\frac{2v}{1 - v^2} dv = \frac{dx}{x}$.Integrating both sides: $-\ln|1 - v^2| = \ln|x| + \ln|c|$.$-\ln|1 - v^2| = \ln|cx| \implies \ln|1 - v^2|^{-1} = \ln|cx| \implies \frac{1}{1 - v^2} = cx$.Since $v = \frac{y}{x}$,we have $\frac{1}{1 - (y/x)^2} = cx \implies \frac{x^2}{x^2 - y^2} = cx \implies x = c(x^2 - y^2)$.

The solution of the differential equation $(x^2 + y^2)dx = 2xydy$ is

Similar Questions

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

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If $y=y(x)$ is the solution of $x \frac{d y}{d x}=y+x e^{-\left(\frac{y}{x}\right)}$ with $y(1)=\log e$,then find $y(e)$.

Show that the differential equation is a homogeneous equation and find its solution:
$\left\{x \cos \left(\frac{y}{x}\right)+y \sin \left(\frac{y}{x}\right)\right\} y \, dx = \left\{y \sin \left(\frac{y}{x}\right)-x \cos \left(\frac{y}{x}\right)\right\} x \, dy$

Consider the differential equation $\frac{dy}{dx} = \frac{y^3}{2(xy^2 - x^2)}$.
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