Question

Y= ln (x+y), y" - как решить? подскажите

Answer 1

Скан решения задания

Answer 2

$\displaystyle y(x) = \ln(x+y(x))\\\\ y'(x) = \frac{1+y'(x)}{x+y(x)} = \frac{1+y'(x)}{e^{y(x)}}\\\\ y'(x) = \frac{1}{e^{y(x)}-1}\\\\ y''(x) = -\frac{y'(x)e^{y(x)}}{(e^{y(x)}-1)^2} = -\frac{x+y(x)}{(x+y(x)-1)^3}$