We #160;have,dydx=yxlog #160;yx+1This #160;is #160;a #160;homogeneous #160;differential #160;equation.Putting #160;y=vx #160;and #160;dydx=v ...

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Equation of a Plane Parallel to a Given Plane

dydx = yxlogy...

Question

$\frac{d y}{d x} = \frac{y}{x} \{\log (\frac{y}{x}) + 1\}$

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Solution

$We have, \frac{d y}{d x} = \frac{y}{x} \{\log (\frac{y}{x}) + 1\} This is a homogeneous differential equation . Putting y = v x and \frac{d y}{d x} = v + x \frac{d v}{d x}, we get v + x \frac{d v}{d x} = v \{\log v + 1\} \Rightarrow x \frac{d v}{d x} = v \log v + v - v \Rightarrow \frac{1}{v \log v} d v = \frac{1}{x} d x Integrating both sides, we get \int \frac{1}{v \log v} d v = \int \frac{1}{x} d x Putting \log v = t, we get d v = v d t ∴ \int \frac{1}{v \times t} \times v d t = \int \frac{1}{x} d x \Rightarrow \int \frac{d t}{t} = \int \frac{1}{x} d x \Rightarrow \log |t| = \log |x| + \log C \Rightarrow t = C x . . . . . (1) Substituting the value of t in (1), we get \log v = C x Putting v = \frac{y}{x}, we get \Rightarrow \log (\frac{y}{x}) = C x Hence, \log (\frac{y}{x}) = C x is the required solution .$

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