If intlimits_a^x {t,y(t)dt} = x^2 + y(x),then | vedclass

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

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$y = 2 - (2 + a^2)e^{\frac{x^2 - a^2}{2}}$

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(A) Given the equation $\int\limits_a^x {t\,y(t)dt} = x^2 + y(x)$.Differentiating both sides with respect to $x$ using the Leibniz integral rule:$x\,y(x) = 2x + y'(x)$Rearranging the terms to form a linear differential equation:$y'(x) + x\,y(x) = 2x$This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = x$ and $Q(x) = 2x$.The integrating factor $(I.F.)$ is:$I.F. = e^{\int x\,dx} = e^{\frac{x^2}{2}}$Multiplying both sides by the $I.F.$:$e^{\frac{x^2}{2}} \frac{dy}{dx} + x\,e^{\frac{x^2}{2}} y = 2x\,e^{\frac{x^2}{2}}$$\frac{d}{dx} \left( y\,e^{\frac{x^2}{2}} \right) = 2x\,e^{\frac{x^2}{2}}$Integrating both sides with respect to $x$:$y\,e^{\frac{x^2}{2}} = \int 2x\,e^{\frac{x^2}{2}} dx + C$Let $u = \frac{x^2}{2}$,then $du = x\,dx$:$y\,e^{\frac{x^2}{2}} = 2\,e^{\frac{x^2}{2}} + C$Dividing by $e^{\frac{x^2}{2}}$:$y = 2 + C\,e^{-\frac{x^2}{2}}$To find $C$,use the initial condition at $x = a$:$\int\limits_a^a {t\,y(t)dt} = a^2 + y(a) \implies 0 = a^2 + y(a) \implies y(a) = -a^2$Substituting $x = a$ into the general solution:$-a^2 = 2 + C\,e^{-\frac{a^2}{2}}$$C\,e^{-\frac{a^2}{2}} = -(2 + a^2)$$C = -(2 + a^2)e^{\frac{a^2}{2}}$Substituting $C$ back into the general solution:$y = 2 - (2 + a^2)e^{\frac{a^2}{2}} e^{-\frac{x^2}{2}}$$y = 2 - (2 + a^2)e^{\frac{a^2 - x^2}{2}}$Note: The provided options use $e^{\frac{x^2 - a^2}{2}}$. Re-evaluating the sign of the derivative: $x\,y(x) = 2x + y'(x) \implies y' - x\,y = -2x$. $I.F. = e^{-\frac{x^2}{2}}$. $y\,e^{-\frac{x^2}{2}} = \int -2x\,e^{-\frac{x^2}{2}} dx = 2\,e^{-\frac{x^2}{2}} + C$. $y = 2 + C\,e^{\frac{x^2}{2}}$. At $x=a, y=-a^2 \implies -a^2 = 2 + C\,e^{\frac{a^2}{2}} \implies C = -(2+a^2)e^{-\frac{a^2}{2}}$. Thus $y = 2 - (2+a^2)e^{\frac{x^2-a^2}{2}}$.

If $\int\limits_a^x {t\,y(t)dt} = x^2 + y(x)$,then $y$ as a function of $x$ is

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