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Separable Differentiable
Equations
Alternatively, we can find the solution to the DE in example 2 from yesterday
by rearranging the equation first, y >0.
dy
dx
 2y
1
dy  2dx
y
The variable in the separated differential
indicates how to take the antiderivative
Note: this is where the
‘+C’ appears
ln y  2x  C
y  e ( 2x c )
y  e 2x e C
y  C 1e 2x
The constant C1 is the
simplification of eC
Alternatively, we can find the solution to the DE in example 2 from yesterday
by rearranging the equation first in another way, y >0.
dy
dx
 2y
dx
1

dy  2y
1
x
ln y  C
2
2x  2C  ln y
y  e2x2C
y  e( 2x 2C)
y  e2x e2C
y  C 1e 2x
Take the reciprocals of both
sides
Solve the DE where
dy
dx
dy
dx
 2y  4 , y > 2
 2(y  2)
1
dy  2dx
y2
ln( y  2)  2x  c
y  2  e 2x C
y  C1e 2x  2
Factor out the
leading coefficient
Alternatively:
Solve the DE where
dy
dx
dy
dx
 2y  4 , y > 2
 2(y  2)
d( y  2)
 2( y  2)
dx
Ok this is different…
But consider the following:
d 2
( x )  2x
dx
and
From our
pattern of
slide 1
dN
 2N
dx
N  Ce  2 x
( y  2)  Ce2x
y  Ce2x  2
d 2
( x  5)  2x
dx
Thus
d 2
d 2
(x ) 
( x  5)
dx
dx