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

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(A) Given the equation: $\int_{a}^{x} t y(t) dt = x^2 + y(x)$  $(1)$Differentiating both sides with respect to $x$ using Leibniz's rule:$x y(x) = 2x +

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

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(A) Given the equation: $\int_{a}^{x} t y(t) dt = x^2 + y(x)$  $(1)$Differentiating both sides with respect to $x$ using Leibniz's rule:$x y(x) = 2x + \frac{dy}{dx}$Rearranging the terms:$\frac{dy}{dx} = x y - 2x = x(y - 2)$Separating the variables:$\int \frac{dy}{y - 2} = \int x dx$Integrating both sides:$\ln |y - 2| = \frac{x^2}{2} + C$$y - 2 = K e^{\frac{x^2}{2}}$ (where $K = \pm e^C$)At $x = a$,the integral $\int_{a}^{a} t y(t) dt = 0$,so from $(1)$:$0 = a^2 + y(a) \Rightarrow y(a) = -a^2$Substituting $x = a$ and $y = -a^2$ into the general solution:$-a^2 - 2 = K e^{\frac{a^2}{2}}$$K = -(a^2 + 2) e^{-\frac{a^2}{2}}$Substituting $K$ back into the equation:$y - 2 = -(a^2 + 2) e^{-\frac{a^2}{2}} \cdot e^{\frac{x^2}{2}}$$y - 2 = -(a^2 + 2) e^{\frac{x^2 - a^2}{2}}$$y = 2 - (2 + a^2) e^{\frac{x^2 - a^2}{2}}$

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

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