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(C) $x(x^{2}-1) \frac{dy}{dx} = 1$$\Rightarrow dy = \frac{dx}{x(x-1)(x+1)}$Integrating both sides:$\int dy = \int \frac{1}{x(x-1)(x+1)} dx \quad \dots

Vedclass · Accepted Answer

$y=\frac{1}{2} \log \left| \frac{4(x^{2}-1)}{3x^{2}} \right|$

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(C) $x(x^{2}-1) \frac{dy}{dx} = 1$$\Rightarrow dy = \frac{dx}{x(x-1)(x+1)}$Integrating both sides:$\int dy = \int \frac{1}{x(x-1)(x+1)} dx \quad \dots(1)$Using partial fractions:$\frac{1}{x(x-1)(x+1)} = \frac{A}{x} + \frac{B}{x-1} + \frac{C}{x+1}$$1 = A(x-1)(x+1) + Bx(x+1) + Cx(x-1)$For $x=0, 1 = A(-1)(1) \Rightarrow A = -1$For $x=1, 1 = B(1)(2) \Rightarrow B = \frac{1}{2}$For $x=-1, 1 = C(-1)(-2) \Rightarrow C = \frac{1}{2}$Substituting back into $(1)$:$y = -\int \frac{1}{x} dx + \frac{1}{2} \int \frac{1}{x-1} dx + \frac{1}{2} \int \frac{1}{x+1} dx$$y = -\log |x| + \frac{1}{2} \log |x-1| + \frac{1}{2} \log |x+1| + C_1$$y = \frac{1}{2} \log \left| \frac{(x-1)(x+1)}{x^2} \right| + C_1 = \frac{1}{2} \log \left| \frac{x^2-1}{x^2} \right| + C_1$Given $y=0$ when $x=2$:$0 = \frac{1}{2} \log \left| \frac{4-1}{4} \right| + C_1 \Rightarrow 0 = \frac{1}{2} \log \left( \frac{3}{4} \right) + C_1 \Rightarrow C_1 = -\frac{1}{2} \log \left( \frac{3}{4} \right) = \frac{1}{2} \log \left( \frac{4}{3} \right)$Thus,$y = \frac{1}{2} \log \left| \frac{x^2-1}{x^2} \right| + \frac{1}{2} \log \left( \frac{4}{3} \right) = \frac{1}{2} \log \left| \frac{4(x^2-1)}{3x^2} \right|$

Find a particular solution satisfying the given condition:
$x(x^{2}-1) \frac{dy}{dx}=1; y=0$ when $x=2$

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Find a particular solution satisfying the given condition:$x(x^{2}-1) \frac{dy}{dx}=1; y=0$ when $x=2$