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$$y(x) = 2 + \int_8^x (t-ty(t))dt$$ I am having a very hard time doing this problem. (i) Solve the separable differential equation $$y'(x) = x − xy(x)$$ to get $$y(x) = 1 + c \cdot e^{−x^2/2}$$ (ii) Using your answer to part (i), solve the integral equation.

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$$y(x) = 2 + \int_8^x (t-t y(t)) dt \,\, (\clubsuit)$$ Differentiating with respect to $x$, we get that $$y'(x) = x - x y(x) \implies y'(x) + x \cdot y(x) =x \implies y'(x) \cdot e^{x^2/2} + x \cdot e^{x^2/2} \cdot y(x) = x \cdot e^{x^2/2}$$ Hence, we get that $$\dfrac{d \left(y(x) \cdot e^{x^2/2}\right)}{dx} = \dfrac{d (e^{x^2/2})}{dx}$$ Hence, $$y(x) = 1 + c \cdot e^{-x^2/2} \,\, (\spadesuit)$$ To find $c$, plug in $(\spadesuit)$ into $(\clubsuit)$.

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  • $\begingroup$ so would c=2? or 8 since y(8)=2? or should i just completely disregard the 8 because of the FTC? $\endgroup$ – junebug Feb 26 '13 at 1:42

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