The integrating factor of the first order dif | vedclass

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(B) Given the differential equation: $x^{2}(x^{2}-1) \frac{dy}{dx} + x(x^{2}+1)y = x^{2}-1$.Divide throughout by $x^{2}(x^{2}-1)$ to get it in the sta

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$x-\frac{1}{x}$

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(B) Given the differential equation: $x^{2}(x^{2}-1) \frac{dy}{dx} + x(x^{2}+1)y = x^{2}-1$.Divide throughout by $x^{2}(x^{2}-1)$ to get it in the standard form $\frac{dy}{dx} + Py = Q$:$\frac{dy}{dx} + \frac{x^{2}+1}{x(x^{2}-1)}y = \frac{1}{x^{2}}$.Here, $P = \frac{x^{2}+1}{x(x^{2}-1)}$.The integrating factor $(IF)$ is given by $e^{\int P dx}$.$\int P dx = \int \frac{x^{2}+1}{x(x-1)(x+1)} dx$.Using partial fractions: $\frac{x^{2}+1}{x(x-1)(x+1)} = \frac{A}{x} + \frac{B}{x-1} + \frac{C}{x+1}$.Solving for constants: $x^{2}+1 = A(x^{2}-1) + Bx(x+1) + Cx(x-1)$.For $x=0$, $1 = -A \Rightarrow A = -1$.For $x=1$, $2 = 2B \Rightarrow B = 1$.For $x=-1$, $2 = 2C \Rightarrow C = 1$.So, $\int P dx = \int (\frac{-1}{x} + \frac{1}{x-1} + \frac{1}{x+1}) dx = -\ln|x| + \ln|x-1| + \ln|x+1| = \ln|\frac{x^{2}-1}{x}|$.$IF = e^{\ln|\frac{x^{2}-1}{x}|} = \frac{x^{2}-1}{x} = x - \frac{1}{x}$.

The integrating factor of the first order differential equation $x^{2}(x^{2}-1) \frac{dy}{dx} + x(x^{2}+1)y = x^{2}-1$ is

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