The solution of the differential equation 2x | vedclass

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(C) Given the differential equation: $2x \frac{dy}{dx} - y = 0$. Rearranging the terms,we get: $2x \frac{dy}{dx} = y$. Separating the variables: $\fra

Vedclass · Accepted Answer

$A$ parabola

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(C) Given the differential equation: $2x \frac{dy}{dx} - y = 0$. Rearranging the terms,we get: $2x \frac{dy}{dx} = y$. Separating the variables: $\frac{dy}{y} = \frac{dx}{2x}$. Integrating both sides: $\int \frac{dy}{y} = \frac{1}{2} \int \frac{dx}{x}$. This gives: $\ln|y| = \frac{1}{2} \ln|x| + C$. Using the condition $y(1) = 2$: $\ln(2) = \frac{1}{2} \ln(1) + C \implies C = \ln(2)$. Substituting $C$ back: $\ln(y) = \ln(\sqrt{x}) + \ln(2) = \ln(2\sqrt{x})$. Thus,$y = 2\sqrt{x}$,which implies $y^2 = 4x$. The equation $y^2 = 4ax$ represents a parabola. Therefore,the correct option is $C$.

The solution of the differential equation $2x \frac{dy}{dx} - y = 0$ with the condition $y(1) = 2$ represents . . . . . . .

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