Question

help! yy' - ex = 0 , and y = 4 when x = 0, which means that: (see attachment)

Answer 1

\[y \frac{dy}{dx}-e^x=0\]\[y \frac{dy}{dx}=e^x\]\[ydy=e^xdx\]integrate both sides\[\frac{y^2}{2}=e^x+C\]
y=4 when x=0\[\frac{4^2}{2}=e^0+C\]\[8=1+C\]
C=7

so \[y^2=2e^x+14\] which is D

Answer 2

i got it right. thanks!

Answer 3

:)