The solution of the differential equation x^2 | vedclass

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(C) Given differential equation: $x^2 \frac{dy}{dx} - xy = 1$Divide by $x^2$: $\frac{dy}{dx} - \frac{1}{x} y = \frac{1}{x^2}$This is a linear differen

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$2xy - 2cx^2 + 1 = 0$

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(C) Given differential equation: $x^2 \frac{dy}{dx} - xy = 1$Divide by $x^2$: $\frac{dy}{dx} - \frac{1}{x} y = \frac{1}{x^2}$This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$, where $P = -\frac{1}{x}$ and $Q = \frac{1}{x^2}$.Integrating factor $IF = e^{\int P dx} = e^{\int -\frac{1}{x} dx} = e^{-\ln|x|} = \frac{1}{x}$.The solution is $y(IF) = \int Q(IF) dx + c$.$y(\frac{1}{x}) = \int \frac{1}{x^2} \cdot \frac{1}{x} dx + c = \int x^{-3} dx + c$.$\frac{y}{x} = \frac{x^{-2}}{-2} + c = -\frac{1}{2x^2} + c$.Multiply by $2x^2$: $2xy = -1 + 2cx^2$.Rearranging gives: $2xy - 2cx^2 + 1 = 0$.

The solution of the differential equation $x^2 \frac{dy}{dx} - xy = 1$ is...

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