STOP AND THINK!
The next step is one where many beginners slip up. To solve forywe ‘take the expo-
nential’ of both sides (131
➤
):
y=elny=ex^2
2 +CNOT
‘y=ex^2(^2) +eC’
We now use (129
➤
)–AGAIN BE CAREFUL, MANY BEGINNERS MAKE
MISTAKES HERE TOO.
eA+B=eAeB
to obtain
y=e
x^2
(^2) eC=Ae
x^2
2
where we have replacedeCbyA, another (positive) arbitrary constant.
Given thaty=1whenx=0wehave
1 =Ae^0 =A
so the required solution is (again, check it in the DE)
y=ex
(^2) / 2
(ii)
dy
dx
=e^2 x+y=e^2 xeyand so, separating the variables
e−ydy=e^2 xdx
Integrating both sides gives
∫
e−ydy=
∫
e^2 xdx+C
or
−e−y=
e^2 x
2
+C
We might as well findCnow
−e^0 =
e^0
2
+C