The equation of the curve which passes throug | vedclass

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(D) Given the slope of the tangent is $\frac{dy}{dx} = 1 + \frac{y}{x} + \left(\frac{y}{x}\right)^2$. This is a homogeneous differential equation. Let

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$	an ^{-1}\left(\frac{y}{x}ight)=\log |x|$

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(D) Given the slope of the tangent is $\frac{dy}{dx} = 1 + \frac{y}{x} + \left(\frac{y}{x}ight)^2$. This is a homogeneous differential equation. Let $y = ux$,then $\frac{dy}{dx} = u + x\frac{du}{dx}$. Substituting this into the equation: $u + x\frac{du}{dx} = 1 + u + u^2$. Subtracting $u$ from both sides: $x\frac{du}{dx} = 1 + u^2$. Separating the variables: $\int \frac{du}{1 + u^2} = \int \frac{dx}{x}$. Integrating both sides: $	an^{-1}(u) = \log|x| + C$. Substituting $u = \frac{y}{x}$: $	an^{-1}\left(\frac{y}{x}ight) = \log|x| + C$. Since the curve passes through $(1,0)$,we have $	an^{-1}(0) = \log|1| + C$,which gives $0 = 0 + C$,so $C = 0$. Thus,the equation is $	an^{-1}\left(\frac{y}{x}ight) = \log|x|$.

The equation of the curve which passes through point $(1,0)$ and has a tangent with slope $1+\frac{y}{x}+\left(\frac{y}{x}\right)^{2}$ is

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