if dy/dx= (1+lnx)y and if y=1 when x=1, then y=

1 answer

dy/dx= (1+lnx)y --->

dy/y = (1 + ln(x)) dx --->

ln(y) = x ln(x) + c

Boundary condition:

y=1 when x=1 ---> c = 0

y = exp(x ln(x)) = x^x