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Equation of Tangent at a Point (x,y) in Terms of f'(x)

If y=yx is th...

Question

If $y = y (x)$ is the solution of $x \int 0 t y d t = x^{2} + y^{2}$ and $y (0) = 1,$ where point $(a, 3)$ satisfies the curve $y = y (x),$ then the value of $a^{2} =$

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Solution

Given:
$x \int 0 t y d t = x^{2} + y^{2}$
differentiating both sides w.r.t. $x,$
$\Rightarrow x y = 2 x + 2 y \frac{d y}{d x} \Rightarrow \frac{2 y}{y - 2} d y = x d x$
integrating both sides,
$\Rightarrow \int (2 + \frac{4}{y - 2}) d y = \int x d x \Rightarrow 2 y + 4 ln | y - 2 | = \frac{x^{2}}{2} + C \Rightarrow x^{2} + c = 4 (y + 2 ln | y - 2 |)$
As, $y (0) = 1; c = 4$
Now, point $(a, 3)$ lies on this curve
$\Rightarrow a^{2} + 4 = 4 (3 + 0) \Rightarrow a^{2} = 8$

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