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(B) Given the family of curves $y=ax+\frac{1}{a}$.Differentiating with respect to $x$,we get $\frac{dy}{dx}=a$.Substituting $a=\frac{dy}{dx}$ into the

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$y^2=3x$

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(B) Given the family of curves $y=ax+\frac{1}{a}$.Differentiating with respect to $x$,we get $\frac{dy}{dx}=a$.Substituting $a=\frac{dy}{dx}$ into the original equation,we get $y=x\left(\frac{dy}{dx}\right)+\frac{1}{\frac{dy}{dx}}$.Multiplying by $\frac{dy}{dx}$,we get $y\left(\frac{dy}{dx}\right)=x\left(\frac{dy}{dx}\right)^2+1$,which is $x\left(\frac{dy}{dx}\right)^2-y\left(\frac{dy}{dx}\right)+1=0$.The order $m$ of this differential equation is $1$ and the degree $r$ is $2$.Thus,$r+m = 2+1 = 3$.The given differential equation is $\frac{dy}{dx}=\frac{y}{2x}$.Separating variables,we get $\frac{dy}{y}=\frac{dx}{2x}$.Integrating both sides,$\ln|y|=\frac{1}{2}\ln|x|+C$,which implies $y^2=kx$.Using the condition $y(1)=\sqrt{r+m}=\sqrt{3}$,we get $(\sqrt{3})^2=k(1)$,so $k=3$.Therefore,the solution is $y^2=3x$.

If the degree of the differential equation corresponding to the family of curves $y=ax+\frac{1}{a}$ (where $a \neq 0$ is an arbitrary constant) is $r$ and its order is $m$,then the solution of $\frac{dy}{dx}=\frac{y}{2x}, y(1)=\sqrt{r+m}$ is

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