The general solution of x dy - y dx = y dy is | vedclass

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(A) Given equation: $x dy - y dx = y dy$ Divide both sides by $xy$: $\frac{x dy - y dx}{xy} = \frac{y dy}{xy} = \frac{dy}{x}$ This does not simplify e

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$y = A e^{-x/y}$

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(A) Given equation: $x dy - y dx = y dy$ Divide both sides by $xy$: $\frac{x dy - y dx}{xy} = \frac{y dy}{xy} = \frac{dy}{x}$ This does not simplify easily. Let us rearrange the original equation: $x dy - y dx = y dy$ $\Rightarrow x dy - y dy = y dx$ $\Rightarrow (x - y) dy = y dx$ $\Rightarrow \frac{dy}{dx} = \frac{y}{x - y}$ This is a homogeneous differential equation. Let $y = vx$,then $dy = v dx + x dv$. $v dx + x dv = \frac{vx}{x - vx} dx = \frac{v}{1 - v} dx$ $x dv = (\frac{v}{1 - v} - v) dx = (\frac{v - v + v^2}{1 - v}) dx = \frac{v^2}{1 - v} dx$ $\frac{1 - v}{v^2} dv = \frac{dx}{x}$ $\int (v^{-2} - v^{-1}) dv = \int \frac{dx}{x}$ $-v^{-1} - \ln|v| = \ln|x| + C$ $-\frac{1}{v} = \ln|vx| + C$ Since $y = vx$,we have $v = \frac{y}{x}$ and $vx = y$: $-\frac{x}{y} = \ln|y| + C$ $\ln|y| = -\frac{x}{y} - C$ $y = e^{-x/y - C} = A e^{-x/y}$ where $A = e^{-C}$.

The general solution of $x dy - y dx = y dy$ is

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