The equation of the curve which passes throug | vedclass

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Question

(A) Given the slope of the curve is $\frac{dy}{dx} = \frac{2y}{x}$.Separating the variables,we get $\frac{dy}{y} = \frac{2dx}{x}$.Integrating both sid

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$y = x^2$

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(A) Given the slope of the curve is $\frac{dy}{dx} = \frac{2y}{x}$.Separating the variables,we get $\frac{dy}{y} = \frac{2dx}{x}$.Integrating both sides,we have $\int \frac{dy}{y} = 2 \int \frac{dx}{x}$.This gives $\ln|y| = 2 \ln|x| + C$,which simplifies to $\ln|y| = \ln|x^2| + C$.Taking the exponential of both sides,we get $y = c x^2$.Since the curve passes through the point $(1, 1)$,we substitute $x = 1$ and $y = 1$ into the equation:$1 = c(1)^2$,which implies $c = 1$.Therefore,the equation of the curve is $y = x^2$.

The equation of the curve which passes through the point $(1, 1)$ and whose slope is given by $\frac{2y}{x}$ is:

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