The solution of (x + log y)dy + ydx = 0 when y(0) = 1 is:
Answer & explanation
Correct answer: option 3
$\int(xdy+ydx)+\int\log ydy=0⇒\int d(xy)+\int\log y dy = c$
⇒ xy + y log y - y = c; y(0) - 1 ⇒ c = -1 ⇒ xy + y log y - y + 1 = 0
The solution of (x + log y)dy + ydx = 0 when y(0) = 1 is:
Correct answer: option 3
$\int(xdy+ydx)+\int\log ydy=0⇒\int d(xy)+\int\log y dy = c$
⇒ xy + y log y - y = c; y(0) - 1 ⇒ c = -1 ⇒ xy + y log y - y + 1 = 0