Question

Solve: y dx - x dy + log x dx = 0

Answer 1

Solution:

1) Since the equation can be converted into standard linear equation by some manipulation

$y dx - x dy + log x dx = 0 \\ \\ \frac{y \: dx}{x \: dx} - \frac{xdy}{x \: dx} + \frac{log \: x \: dx}{x \: dx} = 0 \\ \\ \frac{y}{x} - \frac{dy}{dx} + \frac{log \: x}{x} = 0 \\ \\ \frac{dy}{dx} - \frac{y}{x} = \frac{log \: x}{x}$

compare this equation with standard equation I.e.

$\frac{dy}{dx} + Py = Q\\ \\ here \\ \\ P = \frac{ - 1}{x} \\ \\ Q = \frac{log \: x}{x}$
solution

$I.F.= {e}^{\int\:Pdx} \\ \\ {e}^{\int\frac{- 1}{x}dx} = {e}^{- log \: x} \\ \\ = {e}^{ {log \: x}^{- 1} } \\ \\ = {e}^{log \: \frac{1}{x} } \\ \\ = \frac{1}{x} ...eq1$

Final solution:

$y×I.F. = Q \times I.F. \: dx \\ \\ y \times \frac{1}{x} = \int\frac{log \: x}{x} \times \frac{1}{x} dx$

here integration by parts has to be done

$\int\:\frac{log\:x}{x^{2}}dx\\\\=log\:x\int\:\frac{1}{x^{2}}-\int(\frac{dlog x}{dx}\int\frac{1}{x^{2}}dx)dx\\\\=\frac{-log\:x}{x}-\int(\frac{1}{x}(-\frac{1}{x})dx)\\\\=\frac{-log\:x}{x}+\int(\frac{1}{x^{2}}dx)\\\\=\frac{-log\:x}{x}-\frac{1}{x}+C$

So finally complete solution is

$\frac{y}{x}=\frac{-log\:x}{x}-\frac{1}{x}+C$