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

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(D) Given the differential equation: $x \frac{dy}{dx} = y - x^2$. Dividing by $x$ (assuming $x 
eq 0$): $\frac{dy}{dx} = \frac{y}{x} - x$. This is a

Vedclass · Accepted Answer

$\frac{1}{6}$

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(D) Given the differential equation: $x \frac{dy}{dx} = y - x^2$. Dividing by $x$ (assuming $x 
eq 0$): $\frac{dy}{dx} = \frac{y}{x} - x$. This is a linear differential equation of the form $\frac{dy}{dx} - \frac{1}{x}y = -x$. The integrating factor $IF = e^{\int -\frac{1}{x} dx} = e^{-\ln x} = \frac{1}{x}$. Multiplying by $IF$: $\frac{1}{x} \frac{dy}{dx} - \frac{1}{x^2} y = -1$. Integrating both sides: $\frac{y}{x} = \int -1 dx = -x + c$. So,$y = -x^2 + cx$. Using the initial condition $y(1) = 0$: $0 = -1 + c \implies c = 1$. The curve is $y = -x^2 + x$. The curve intersects the $x$-axis where $y=0$,i.e.,$-x^2 + x = 0 \implies x(1-x) = 0$,so $x=0$ and $x=1$. The area bounded by the curve and the $x$-axis is $\int_0^1 (-x^2 + x) dx$. $= [-\frac{x^3}{3} + \frac{x^2}{2}]_0^1 = -\frac{1}{3} + \frac{1}{2} = \frac{1}{6}$.

$A$ 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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