A curve passing through (2, 3) and satisfying | vedclass

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(D) Given the integral equation: $\int\limits_0^x {t\,y(t)\,dt} = x^2y(x)$.Differentiating both sides with respect to $x$ using the Leibniz rule:$x\,y

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$xy = 6$

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(D) Given the integral equation: $\int\limits_0^x {t\,y(t)\,dt} = x^2y(x)$.Differentiating both sides with respect to $x$ using the Leibniz rule:$x\,y(x) = \frac{d}{dx}(x^2y(x))$$x\,y(x) = x^2y'(x) + 2x\,y(x)$.Rearranging the terms:$x^2y'(x) + x\,y(x) = 0$.Dividing by $x$ (since $x > 0$):$x\,y'(x) + y(x) = 0$.This is a separable differential equation:$x\frac{dy}{dx} = -y$$\frac{dy}{y} = -\frac{dx}{x}$.Integrating both sides:$\ln|y| = -\ln|x| + C$$\ln|y| + \ln|x| = C$$\ln|xy| = C$$xy = k$,where $k = e^C$.Since the curve passes through $(2, 3)$:$(2)(3) = k \implies k = 6$.Thus,the equation of the curve is $xy = 6$.

$A$ curve passing through $(2, 3)$ and satisfying the differential equation $\int\limits_0^x {t\,y(t)\,dt} = x^2y(x)$ for $x > 0$ is

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