If a curve y=f(x), passing through the point | vedclass

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(B) Given differential equation: $2 x^{2} dy = (2 xy + y^{2}) dx$$\frac{dy}{dx} = \frac{2xy + y^{2}}{2x^{2}}$This is a homogeneous differential equati

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$\frac{1}{1+\log _{e} 2}$

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(B) Given differential equation: $2 x^{2} dy = (2 xy + y^{2}) dx$$\frac{dy}{dx} = \frac{2xy + y^{2}}{2x^{2}}$This is a homogeneous differential equation. Let $y = vx$,then $\frac{dy}{dx} = v + x\frac{dv}{dx}$.Substituting these into the equation:$v + x\frac{dv}{dx} = \frac{2x(vx) + (vx)^{2}}{2x^{2}} = \frac{2x^{2}v + x^{2}v^{2}}{2x^{2}} = v + \frac{v^{2}}{2}$$x\frac{dv}{dx} = \frac{v^{2}}{2}$Separating variables:$\frac{2}{v^{2}} dv = \frac{1}{x} dx$Integrating both sides:$\int 2v^{-2} dv = \int \frac{1}{x} dx$$-2v^{-1} = \ln|x| + C$$-\frac{2}{v} = \ln|x| + C$Since $v = \frac{y}{x}$,we have $-\frac{2x}{y} = \ln|x| + C$.The curve passes through $(1, 2)$,so substitute $x=1, y=2$:$-\frac{2(1)}{2} = \ln(1) + C \Rightarrow -1 = 0 + C \Rightarrow C = -1$.Thus,$-\frac{2x}{y} = \ln|x| - 1$,which implies $\frac{2x}{y} = 1 - \ln x$.$y = \frac{2x}{1 - \ln x} \Rightarrow f(x) = \frac{2x}{1 - \ln x}$.Now,calculate $f\left(\frac{1}{2}\right)$:$f\left(\frac{1}{2}\right) = \frac{2(\frac{1}{2})}{1 - \ln(\frac{1}{2})} = \frac{1}{1 - (\ln 1 - \ln 2)} = \frac{1}{1 - (0 - \ln 2)} = \frac{1}{1 + \ln 2}$.

If a curve $y=f(x),$ passing through the point $(1,2),$ is the solution of the differential equation $2 x^{2} dy=\left(2 xy+y^{2}\right) dx,$ then $f\left(\frac{1}{2}\right)$ is equal to

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