The general solution of the differential equa | vedclass

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Question

(C) Given the differential equation: $\frac{dy}{dx} + \frac{y \ln y}{x} = \frac{y(\ln y)^2}{x^2}$.Divide both sides by $y(\ln y)^2$: $\frac{1}{y(\ln y

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$\frac{1}{\ln y} = \frac{1}{2x} + Cx$

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(C) Given the differential equation: $\frac{dy}{dx} + \frac{y \ln y}{x} = \frac{y(\ln y)^2}{x^2}$.Divide both sides by $y(\ln y)^2$: $\frac{1}{y(\ln y)^2} \frac{dy}{dx} + \frac{1}{x \ln y} = \frac{1}{x^2}$.Let $t = \frac{1}{\ln y}$. Then $\frac{dt}{dx} = -\frac{1}{y(\ln y)^2} \frac{dy}{dx}$.Substituting this into the equation,we get: $-\frac{dt}{dx} + \frac{t}{x} = \frac{1}{x^2}$,which simplifies to $\frac{dt}{dx} - \frac{t}{x} = -\frac{1}{x^2}$.This is a linear differential equation of the form $\frac{dt}{dx} + P(x)t = Q(x)$,where $P(x) = -\frac{1}{x}$ and $Q(x) = -\frac{1}{x^2}$.The integrating factor $IF = e^{\int P(x) dx} = e^{-\int \frac{1}{x} dx} = e^{-\ln x} = \frac{1}{x}$.The solution is $t \cdot IF = \int Q(x) \cdot IF dx + C$.$t \cdot \frac{1}{x} = \int (-\frac{1}{x^2}) \cdot \frac{1}{x} dx + C = -\int x^{-3} dx + C = -(\frac{x^{-2}}{-2}) + C = \frac{1}{2x^2} + C$.Multiplying by $x$,we get $t = \frac{1}{2x} + Cx$.Substituting $t = \frac{1}{\ln y}$,we get $\frac{1}{\ln y} = \frac{1}{2x} + Cx$.

The general solution of the differential equation $\frac{dy}{dx} + \frac{y \ln y}{x} = \frac{y(\ln y)^2}{x^2}$ is (where $C$ is an arbitrary constant):

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