or sincee^0 =1,C=−^32 and the solution can be written
e−y=
3
2
−
e^2 x
2
Note that sincee−ymust be positive, we are here restricted toe^2 x≤3.
Solving forygives
−y=ln
∣∣
∣
∣
3
2
−
e^2 x
2
∣∣
∣
∣
or
y=−ln
∣
∣
∣
∣
2
3
−
e^2 x
2
∣
∣
∣
∣
The form of variables separable equations is rather restrictive. Even such a simple
function asF(x,y)=x+ywouldn’t fit into it. However, there are many types of equation
that may be reduced to variables separable by some kind of substitution. Consider, for
example the equation
dy
dx
=
x+y
x
=F(x,y)
whereF(x,y)is of the form
F(x,y)=f
(y
x
)
Such an equation is said to behomogeneous(notto be confused with later use of this
term). If we change our variables fromx,ytoxandv=
y
x
we havey=xvand so
dy
dx
=v+x
dv
dx
and the equation becomes
dy
dx
=v+x
dv
dx
=f(v)
or
x
dv
dx
=f(v)−v
This isseparableand its solution is
∫
dv
f(v)−v
=
∫
dx
x
+C=lnx+C
After evaluating the integral on the left we can then replacevbyy/xto get the solution
in terms ofxandy.