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MTHE 225
HOMEWORK SET 6
1. Solve the differential equation: y 0 = sin2 (x − y).
2. Solve the differential equation: y 0 = (6x − 2y + 1)2/3 + 3.
3. Solve the differential equation: 5y 0 =
5xy + 18x2 + 2y 2
.
x2
6y 2 + 5x2
,
4. Solve the initial value problem: y =
xy
0
1
y(1) = 1.
SOLUTION
1.
Here we use “direct substitution method”.
dy
dy
du
du
=1−
⇒
=1−
. Now substituting these in the
Let u = x − y. That means
dx
dx
dx
dx
differential equation:
dy
y0 =
= sin2 (x − y)
dx
du
⇒1−
= sin2 u
dx
du
⇒
= 1 − sin2 u
dx
du
= cos2 u
⇒
dx
du
⇒
= dx
cos2 u
⇒ sec2 u du = dx
Z
Z
2
⇒ sec u du =
dx
⇒ tan u = x + C
⇒ tan (x − y) = x + C
(implicit solution)
⇒ x − y = arctan (x + C)
⇒ y = x − arctan (x + C)
(explicit solution)
2.
Here we use “direct substitution method”.
du
dy
dy
1
Let u = 6x − 2y + 1. That means
=6−2
⇒
=
dx
dx
dx
2
substituting these in the differential equation:
y 0 = (6x − 2y + 1)2/3 + 3
1 du
= u2/3 + 3
2 dx
1 du
⇒−
= u2/3
2 dx
⇒3−
2
du
1 du
6−
=3−
. Now
dx
2 dx
⇒
du
= −2 dx
u2/3
⇒ u−2/3 du = −2 dx
Z
Z
−2/3
⇒ u
du = −2 dx
⇒
u1/3
1
3
= −2x + C
1
⇒ u1/3 = (−2x + C)
3
1
⇒ (6x − 2y + 1)1/3 = (−2x + C)
(implicit solution)
3
3
1
⇒ (6x − 2y + 1) = (−2x + C)
3
3
1
⇒ 2y = 6x + 1 − (−2x + C)
3
1
1
⇒ y = 3x + − (−2x + C)3
(explicit solution)
2 54
3.
Notice: 5y 0 =
dy
5xy + 18x2 + 2y 2
5xy + 18x2 + 2y 2
⇒
5
=
x2
dx
x2
⇒ 5x2 dy − (5xy + 18x2 + 2y 2 ) dx = 0
(?)
This differential equation“looks like” either in homogeneous form or in exact form. So let us
first check if it in homogeneous form.
Assuming it is in homogeneous form, we have M (x, y) = 5x2 and N (x, y) = 5xy +18x2 +2y 2 .
Now notice:
M (tx, ty) = 5(tx)2 = t2 (x2 ) = t2 M (x, y)
N (tx, ty) = 5(tx)(ty) + 18(tx)2 + 2(ty)2 = t2 (5xy + 18x2 + 2y 2 ) = t2 N (x, y)
Hence the given differential equation is in homogeneous form.
Let y = ux. That means dy = x du + u dx. Now substituting these in (?):
5x2 (x du + u dx) − (5x(ux) + 18x2 + 2(ux)2 ) dx = 0
⇒ 5(x du + u dx) − (5u + 18 + 2u2 ) dx = 0 (dividing both sides by 2)
⇒ 5x du + 5u dx − 5u dx − 18 dx − 2u2 dx = 0
⇒ 5x du − 18 dx − 2u2 dx = 0
3
⇒ 5x du = (18 + 2u2 ) dx
5
⇒ x du = (9 + u2 ) dx
2
du
2 dx
⇒
=
2
9+u
5 x
Z
Z
du
dx
2
⇒
=
2
9+u
5
x
u 2
1
= ln x + C
⇒ arctan
3
3
5
u 6
⇒ arctan
= ln x + C
3
5
y 6
⇒ arctan
= ln x + C
3x
5
y
6
⇒
= tan
ln x + C
3x
5
6
⇒ y = 3x tan
ln x + C
5
4.
Notice: y 0 =
(implicit solution)
(explicit solution)
6y 2 + 5x2
dy
6y 2 + 5x2
⇒
=
⇒ xy dy − (6y 2 + 5x2 ) dx = 0
xy
dx
xy
(?)
This differential equation“looks like” either in homogeneous form or in exact form. So let us
first check if it in homogeneous form.
Assuming it is in homogeneous form, we have M (x, y) = xy and N (x, y) = −(6y 2 + 5x2 ).
Now notice:
M (tx, ty) = (tx)(ty) = t2 (xy) = t2 M (x, y)
N (tx, ty) = −(6(ty)2 + 5(tx)2 ) = t2 N (x, y)
Hence the given differential equation is in homogeneous form.
Let y = ux. That means dy = x du + u dx. Now substituting these in (?):
(x)(ux)(x du + u dx) − (6(ux)2 + 5x2 ) dx = 0
⇒ u(x du + u dx) − (6u2 + 5) dx = 0
⇒ ux du + u2 dx − 6u2 dx − 5 dx = 0
⇒ ux du − 5u2 dx − 5 dx = 0
⇒ ux du = 5(1 + u2 ) dx
4
u
1
du = dx
2
1+u
x
Z
Z
u
1
dx
⇒
du =
2
1+u
x
1
⇒ ln (1 + u2 ) = ln x+C
(on the left apply v−sub with v = 1+u2 , that is dv = 2u du)
2
1
y2
⇒ ln 1 + 2 = ln x + C
(implicit solution)
2
x
y2
⇒ ln 1 + 2 = 2(ln x + C)
x
2
2
y
⇒ 1 + 2 = e2(ln x+C) = e2 ln x e2C = eln x D = x2 D;
(D = e2C )
x
⇒
⇒ x2 + y 2 = x4 D
⇒ y 2 = x4 D − x2
⇒y=
√
√
x4 D − x2 = x x2 D − 1
But we want y(1) = 1. That
means 1 =
√
value problem is: y = x 2x2 − 1.
(explicit solution)
√
D − 1 ⇒ D = 2. Hence the solution of the initial
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