The equation of the curve that passes through | vedclass

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Question

(A) Given the differential equation: $\frac{dy}{dx} = \frac{-2xy}{x^2 + 1}$.Separating the variables,we get: $\frac{dy}{y} = \frac{-2x}{x^2 + 1} dx$.I

Vedclass · Accepted Answer

$y(x^2 + 1) = 4$

Vedclass · Answer

(A) Given the differential equation: $\frac{dy}{dx} = \frac{-2xy}{x^2 + 1}$.Separating the variables,we get: $\frac{dy}{y} = \frac{-2x}{x^2 + 1} dx$.Integrating both sides: $\int \frac{dy}{y} = -\int \frac{2x}{x^2 + 1} dx$.This gives: $\ln|y| = -\ln(x^2 + 1) + C$.Using the property of logarithms: $\ln|y| + \ln(x^2 + 1) = C$,which simplifies to $\ln|y(x^2 + 1)| = C$.Thus,$y(x^2 + 1) = e^C = k$ (where $k$ is a constant).Since the curve passes through the point $(1, 2)$,substitute $x = 1$ and $y = 2$ into the equation:$2(1^2 + 1) = k \implies 2(2) = k \implies k = 4$.Therefore,the equation of the curve is $y(x^2 + 1) = 4$.

The equation of the curve that passes through the point $(1, 2)$ and satisfies the differential equation $\frac{dy}{dx} = \frac{-2xy}{x^2 + 1}$ is

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