Question

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

Answer 1

means that :
D:y^2=2e^x+14

Answer 2

is D

Answer 3

I used seperation of variables so the DE becomes:
ydy/dx=e^x
ydy=e^xdx
integrate both sides to get
y^2/2+k=e^x+c
where k and c are just constants of the integration
y^2+2k=2e^x+2c
y^2=2e^x+2c-2k
now 2c-2k is also just any constant, so let that be A=2c-2k where A is any constant so we;ll have:
y^2=2e^x+A
now find the value of the constant A, so we know that y=4 if x=0:
4^2=2e^0+A
16=2*1+A
A=16-2
A=14
so our actual solution will be:
y^2=2e^x+14 which is letter D