The solution of the differential equation ydx | vedclass

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Question

(A) Given equation: $ydx - xdy + \log x dx = 0$. Dividing by $x dx$ (assuming $x 
eq 0$),we get: $\frac{y}{x} - \frac{dy}{dx} + \frac{\log x}{x} = 0$

Vedclass · Accepted Answer

$y = cx - (1 + \log x)$

Vedclass · Answer

(A) Given equation: $ydx - xdy + \log x dx = 0$. Dividing by $x dx$ (assuming $x 
eq 0$),we get: $\frac{y}{x} - \frac{dy}{dx} + \frac{\log x}{x} = 0$. Rearranging the terms: $\frac{dy}{dx} - \frac{y}{x} = \frac{\log x}{x}$. This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = -\frac{1}{x}$ and $Q = \frac{\log x}{x}$. The integrating factor $(I.F.)$ is $e^{\int P dx} = e^{\int -\frac{1}{x} dx} = e^{-\log x} = \frac{1}{x}$. The solution is given by $y \cdot (I.F.) = \int Q \cdot (I.F.) dx + c$. Substituting the values: $y \cdot \frac{1}{x} = \int \frac{\log x}{x} \cdot \frac{1}{x} dx + c = \int \frac{\log x}{x^2} dx + c$. Using integration by parts for $\int \frac{\log x}{x^2} dx$: Let $u = \log x$ and $dv = x^{-2} dx$. Then $du = \frac{1}{x} dx$ and $v = -\frac{1}{x}$. $\int \frac{\log x}{x^2} dx = (\log x)(-\frac{1}{x}) - \int (-\frac{1}{x})(\frac{1}{x}) dx = -\frac{\log x}{x} + \int x^{-2} dx = -\frac{\log x}{x} - \frac{1}{x}$. Thus,$\frac{y}{x} = -\frac{\log x}{x} - \frac{1}{x} + c$. Multiplying by $x$,we get $y = -\log x - 1 + cx$,which is $y = cx - (1 + \log x)$.

The solution of the differential equation $ydx - xdy + \log x dx = 0$ is

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