Practicing Success
Practicing Success
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If $x\frac{dy}{dx}=y(\log y-\log x + 1)$ then the solution of the equation is: |
$\log \frac{x}{y}=cy$ $\log \frac{y}{x}=cy$ $\log \frac{x^2}{y}=cy$ $\log \frac{y}{x}=cx$ |
$\log \frac{y}{x}=cx$ |
$x\,dy-y\,dx=y\log(\frac{y}{x})dx⇒\frac{x\,dy-y\,dx}{x^2}=\frac{y}{x}.\log(\frac{y}{x}).\frac{dx}{x}⇒\frac{d(\frac{y}{x})}{(\frac{y}{x})\log(\frac{y}{x})}=\frac{dy}{x}$ Substitute $\log (\frac{y}{x})=t⇒\log (\frac{y}{x})=cx$ |
