x^3 (d^2 y/dx^2 )+x^2 (dy/dx)+xy=2(x^2 −1) y(1)=0 y′((1/2))=5


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Question Number 289 by 123456 last updated on 25/Jan/15

$x^{3} \frac{d^{2} y}{{d x}^{2}} + x^{2} \frac{d y}{d x} + x y = 2 (x^{2} - 1)$ $y (1) = 0$ $y^{'} (\frac{1}{2}) = 5$
	Answered by prakash jain last updated on 18/Dec/14

	$y_{h} = C_{1} \cos (\ln ∣ x ∣) + C_{2} \sin (\ln ∣ x ∣) (s e e Q 271)$ $y_{p} = A x + \frac{B}{x} + C$ $y^{'} = A - \frac{B}{x^{2}}$ $y^{″} = \frac{2 B}{x^{3}}$ $2 B + x^{2} (A - \frac{B}{x^{2}}) + x (A x + \frac{B}{x} + C) = 2 x^{2} - 2$ $2 B + {A x}^{2} - B + {A x}^{2} + B + C x = 2 x^{2} - 2$ $B = - 1, C = 0, A = 1$ $y_{p} = x - \frac{1}{x}$ $y = C_{1} \cos (\ln ∣ x ∣) + C_{2} \sin (\ln ∣ x ∣) + x - \frac{1}{x}$ $y (1) = 0 \Rightarrow C_{1} = 0$ $y^{'} (\frac{1}{2}) = 5$ $y^{'} = \frac{C_{2}}{x} \cos (\ln ∣ x ∣) + 1 + \frac{1}{x^{2}}$ $y^{'} (\frac{1}{2}) = 2 C_{2} \sin (\ln \frac{1}{2}) + 1 + 4 = 5$ $C_{2} = 0$ $y = x - \frac{1}{x}$
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