calculus
posted by an on .
if dy/dx= (1+lnx)y and if y=1 when x=1, then y=

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
posted by an on .
if dy/dx= (1+lnx)y and if y=1 when x=1, then y=
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