A curve satisfying the initial condition,y(1) | vedclass

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(D) The given differential equation is $x \frac{dy}{dx} = y - x^2$,which can be rewritten as $\frac{dy}{dx} - \frac{1}{x}y = -x$.This is a linear diff

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$\frac{1}{6}$

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(D) The given differential equation is $x \frac{dy}{dx} = y - x^2$,which can be rewritten as $\frac{dy}{dx} - \frac{1}{x}y = -x$.This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = -\frac{1}{x}$ and $Q(x) = -x$.The integrating factor is $IF = e^{\int P(x) dx} = e^{-\int \frac{1}{x} dx} = e^{-\ln|x|} = \frac{1}{x}$.The general solution is $y \cdot IF = \int Q(x) \cdot IF dx + C$,so $y \cdot \frac{1}{x} = \int (-x) \cdot \frac{1}{x} dx + C$.$y \cdot \frac{1}{x} = \int -1 dx + C = -x + C$,which gives $y = -x^2 + Cx$.Using the initial condition $y(1) = 0$,we get $0 = -1^2 + C(1)$,so $C = 1$.The equation of the curve is $y = x - x^2$.The curve intersects the $x$-axis where $y = 0$,so $x - x^2 = 0$,which gives $x(1 - x) = 0$,i.e.,$x = 0$ and $x = 1$.The area bounded by the curve and the $x$-axis is $\int_{0}^{1} (x - x^2) dx = [\frac{x^2}{2} - \frac{x^3}{3}]_{0}^{1} = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}$.

$A$ curve satisfying the initial condition,$y(1) = 0,$ satisfies the differential equation,$x \frac{dy}{dx} = y - x^2.$ The area bounded by the curve and the $x$-axis is

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