The solution of the differential equation 2xy | vedclass

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Question

(D) The given differential equation is $2xy \frac{dy}{dx} = x^2 + 3y^2$,which is a homogeneous differential equation.Rewrite it as $\frac{dy}{dx} = \f

Vedclass · Accepted Answer

$x^2 + y^2 = px^3$

Vedclass · Answer

(D) The given differential equation is $2xy \frac{dy}{dx} = x^2 + 3y^2$,which is a homogeneous differential equation.Rewrite it as $\frac{dy}{dx} = \frac{x^2 + 3y^2}{2xy}$.Substitute $y = vx$,so $\frac{dy}{dx} = v + x \frac{dv}{dx}$.Substituting these into the equation: $v + x \frac{dv}{dx} = \frac{x^2 + 3(vx)^2}{2x(vx)} = \frac{x^2(1 + 3v^2)}{2x^2v} = \frac{1 + 3v^2}{2v}$.Then,$x \frac{dv}{dx} = \frac{1 + 3v^2}{2v} - v = \frac{1 + 3v^2 - 2v^2}{2v} = \frac{1 + v^2}{2v}$.Separating variables: $\frac{2v}{1 + v^2} dv = \frac{1}{x} dx$.Integrating both sides: $\int \frac{2v}{1 + v^2} dv = \int \frac{1}{x} dx$.This gives $\ln(1 + v^2) = \ln|x| + \ln|p|$,where $\ln|p|$ is the constant of integration.So,$1 + v^2 = px$,which means $1 + \frac{y^2}{x^2} = px$.Multiplying by $x^2$,we get $x^2 + y^2 = px^3$.

The solution of the differential equation $2xy \frac{dy}{dx} = x^2 + 3y^2$ is (where $p$ is a constant):

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