18.03 Substitution Methods (old topic 12 notes)
Straight substitution
dy
= F (ax + by + c)
dx
Let v = ax + by + c ⇒
dv
dx
dy
= a + b dx
= a + bF (v): separable DE
dy
= (x + 9y + 7)2 .
Example: Solve dx
dy
dv
Let Rv = x + 9yR + 7 ⇒ dx
= 1 + 9 dx
= 1 + 9v 2
dv
dx ⇒ 13 tan−1 3v = x + C ⇒ v = 31 tan(3x + C)
⇒ 1+9v
2 =
⇒ x + 9y + 7 = 31 tan(3x + C) ⇒ y = 19 ( 13 tan(3x + C) − x − 7)
Note: y only defined on an interval,
e.g. if C = 0 then only defined for −π/2 < 3x < π/2.
dy
5
Example: Solve dx
R =dv(x +Ry + 1) ⇒ v = x + y + 1 ⇒
Separate variables: 1+v5 = dx –painful
dv
dx
= 1 + v5.
Homogeneous equations: y 0 = f (y/x)
Example: Solve y 0 =
x3 +3xy 2 +y 3
x2 y+xy 2
3
–same degree top and bottom.
Divide top and bottom by x ⇒ y 0 =
Example: Solve y 0 =
xy
x2 −y 2
2
1+3(y/x)2 +(y/x)3
.
y/x+(y/x)2
(prime means deriv. w.r.t. x).
y/x
Divide top and bottom by x ⇒ y 0 = 1−(y/x)
2
0
0
Let w = y/x ⇒ wx = y ⇒ w x + w = y
w
w3
w
0
Separable –always the case.
⇒ w0 x + w = 1−w
2 ⇒ w x = 1−w 2 − w = 1−w 2
R 1−w2
R 1
1
⇒
dw = x dx ⇒ − 2w2 − ln w = ln x + C (implicit solution)
w3
3/2
1/2 2
x
Example: Solve y 0 = x y+y
= y/x + (y/x)1/2 .
x5/2
R dw
R dx
= x ⇒ 2w1/2 = ln x + C ⇒ w = ( 21 ln x + C)2
w1/2
⇒ y = x( 21 ln x + C)2
dy
Bernoulli Equations: dx
+ p(x)y = q(x)y n
Looks almost like a linear DE. Algebra ⇒ y −n y 0 + py 1−n = q
Substitute w = y 1−n ⇒ w0 = (1 − n)y −n y 0
1
⇒ 1−n
w0 + pw = q –Linear DE
Note: works as long as n 6= 1 –in this case DE is already linear.
Example: Solve x2 y 0 + xy + y 2 = 0 (this is also homogeneous).
y 0 + x1 y = − x12 y 2 ⇒ y −2 y 0 + x1 y −1 = − x12
Let w = y −1 ⇒ w0 = −y −2 y 0
Substituting:
−w0 + x1 w = − x12 ⇒ w0 − x1 w = x1R2 –Linear DE
1
Use the variation of parameters formula: wh = e− − x dx = eln x = x
Z
1
1
−1 + Cx2
⇒ w=x
+
C
=
x
−
+
C
=
1
1
x3
2x2
2x
1
2x
Back substitute: y = =
w
Cx2 − 1
1
18.03 substitution methods
Double Substitution (We won’t do this.)
x+y
dy
= 3+x+2y
.
Example: Solve dx
Let u = x + y, v = 3 + x + 2y
Consider u the independent variable, i.e., v = v(u)
dx
dv
x = 2u − v + 3, y = −u + v − 3 ⇒ du
= 2 − du
,
2−dv/du
dy
dy du
u
⇒ dx = du · dx = −1+dv/du = v
dv
= 2u+v
(homogeneous DE).
du
u+v
dy
du
= −1 +
dv
du
’Flipping’
Example: Solve
dx
1
dt
= 3
= x3 − x2 t –Linear with t = dependent
. ⇒
2
dt
x − tx
dx
variable).
End of substitution methods notes
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