The general solution of x(x-1) frac{dy}{dx} = | vedclass

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(C) Given the differential equation: $x(x-1) \frac{dy}{dx} = x^3(2x-1) + (x-2)y$. Divide by $x(x-1)$ to write it in the standard linear form $\frac{dy

Vedclass · Accepted Answer

$y(x-1) = x^3(x-1) + cx^2$,where $c$ is the constant of integration.

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(C) Given the differential equation: $x(x-1) \frac{dy}{dx} = x^3(2x-1) + (x-2)y$. Divide by $x(x-1)$ to write it in the standard linear form $\frac{dy}{dx} + P(x)y = Q(x)$: $\frac{dy}{dx} - \frac{x-2}{x(x-1)}y = \frac{x^3(2x-1)}{x(x-1)} = \frac{x^2(2x-1)}{x-1}$. Here,$P(x) = -\frac{x-2}{x(x-1)} = -(\frac{2}{x} - \frac{1}{x-1}) = \frac{1}{x-1} - \frac{2}{x}$. The integrating factor $IF = e^{\int P(x) dx} = e^{\int (\frac{1}{x-1} - \frac{2}{x}) dx} = e^{\ln|x-1| - 2\ln|x|} = \frac{x-1}{x^2}$. Multiplying the linear equation by $IF$: $\frac{d}{dx} [y \cdot \frac{x-1}{x^2}] = \frac{x^2(2x-1)}{x-1} \cdot \frac{x-1}{x^2} = 2x-1$. Integrating both sides with respect to $x$: $y \cdot \frac{x-1}{x^2} = \int (2x-1) dx = x^2 - x + c$. $y \cdot \frac{x-1}{x^2} = x(x-1) + c$. Multiplying by $x^2$: $y(x-1) = x^3(x-1) + cx^2$. Thus,the correct option is $C$.

The general solution of $x(x-1) \frac{dy}{dx} = x^3(2x-1) + (x-2)y$ is

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