-
8/9/2019 Eqs. Solvable for y
1/19
EQUATIONSOLVABLE FOR y
-
8/9/2019 Eqs. Solvable for y
2/19
Solve
(1)Soln: Differentiate eq (1) w r t x, we have
(2)
As then eq (2) will become
pxpy +2
dx
dp
dx
dpxpp
dx
dy +22p
dx
dy =dx
dp
dx
dp
xppp +220)12(
2 =dx
dpxppp
-
8/9/2019 Eqs. Solvable for y
3/19
12
2
++=
xp
pp
dx
dp
)1(
12
=
pp
px
dp
dx
)1(
1
1
2 = ppp xdpdx
)1(
1
1
2 ppp xdpdx (3)which is linear in x
So nth IF is ])1exp[ln(1
2exp
2
ppdp
-
8/9/2019 Eqs. Solvable for y
4/19
Multiplying eq (3) by
2)1( pIF
2
)1( p)1(
)1(
1
)1(2)1(
222 + pppppxdpdxp
p
ppx
dp
dxp
)1()1(2)1(
2
ppx
dp
dxp
11)1(2)1(
2 +
-
8/9/2019 Eqs. Solvable for y
5/19
Integrate it
put this value of x in eq (1)
ppx
dp
d 11])1([
2 +cpppx +ln])1([ 2
2)1(
ln
+
= pppc
x
pp
ppcpy ++= ])1( ln[ 22
pppcppy +)ln()1( 22
-
8/9/2019 Eqs. Solvable for y
6/19
Solve
(1)Soln: Differentiate eq (1) w r t x, we have
(2)
As then eq (2) will become
032 =yxpp
dx
dpp
dx
dpxp
dx
dy233 +
pdx
dy =dx
dpp
dx
dpxpp 233 +
0232 =dx
dpp
dx
dpxp
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8/9/2019 Eqs. Solvable for y
7/19
12
3 p
x
dp
dx(3)
which is linear in x
So its IF is ]ln2
3exp[
2
3exp p
p
dp =
=
0)23(2 =dx
dppxp
)23(
2
px
p
dx
dp += p pxdpdx 2 )23( +=
-
8/9/2019 Eqs. Solvable for y
8/19
Multiplying eq (3) by
2
3
pIF =2
3
p
2
3
2
3
2
3
2
3pp
p
x
dp
dxp
2
3
2
1
2
3
2
3pxp
dp
dxp
2
3
2
3
)( pxp
dp
d
-
8/9/2019 Eqs. Solvable for y
9/19
Integrate it
Taking square of both sides
The soln is
and the given equation
1
2
5
2
3
2
5)( c
pxp +
125
23
525 cpxp +12
5
2
3
525 cpxp =2
1
2325)25( cpxp =
12
3
5)25( cpxp =
cpxp =23 )25(
-
8/9/2019 Eqs. Solvable for y
10/19
Solve
(1)Soln: Differentiate eq (1) w r t x, we have
(2)
As then eq (2) will become
09322 =xypxp
xpdx
dpxp
dx
dpy
dx
dyp 18233
2 +p
dx
dy =xpdx
dpxpdx
dpyp 18233 22 +
xp
dx
dpxpy 182)23(
2 +
-
8/9/2019 Eqs. Solvable for y
11/19
pxpdx
dpxpxxp 182)29(
3222 +)9(2)9(
22 xppdx
dpxpx
)9(2)9(22
xppdxdpxpx
xpdx
dpxp
p
xxp182]2
9[
222 +
-
8/9/2019 Eqs. Solvable for y
12/19
pdx
dpx 2
x
p
dx
dp 2=px
dpdx
2=
pdp
xdx
2=
1ln2
1
ln2
1
ln cpx + pcx 1lnln =pcx 1 1
2pcx =
Put the value of p in eq (1)
0932242 =xycxxxc
09332 =ycxc
pcx =2
-
8/9/2019 Eqs. Solvable for y
13/19
Solve
Solve 0)(2 =xpyxyp
0)(22 =xypyxyp
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8/9/2019 Eqs. Solvable for y
14/19
EQUATIONSOLVABLE FOR x
-
8/9/2019 Eqs. Solvable for y
15/19
Solve (1)
Soln: (2)
Differentiate eq (2) w r t y, we have
(3)
As then eq (3) will become
21 pxp +
dydp
dydp
pdydx +
21
dy
dx
p =1
pp
x +1
dy
dp
pp
]1
1[1
2
-
8/9/2019 Eqs. Solvable for y
16/19
By integrating, we get
(3)
Thus eq (2) and eq (3) constitute the solution of eq(1)
dy
dp
p
p ]1
[1 dpp
pdy ]1
[ p
pyc ln
2
2 cppy 2ln22
2
-
8/9/2019 Eqs. Solvable for y
17/19
Solve
Soln:
(1)
Differentiate eq (1) w r t y, we have
+ 21tan ppxp
+
+ = )1()1(
2)1(
222
22
p
dy
dp
p
dy
dp
pdy
dp
p
dy
dx
2
1
1tan
ppxp +
pp
px
1
2
tan1
++
-
8/9/2019 Eqs. Solvable for y
18/19As then eq (2) will becomepdx
dy
=
+
+
= )1()1()21(
222
22
p
dy
dp
p
dy
dppp
dy
dx
dydp
ppp
dydx + += 22
22
)1()11(
dydp
ppp
dydx + += 22
22
)1()11(
-
8/9/2019 Eqs. Solvable for y
19/19
(2)
As then eq (2) will become
and the givenequation
pdx
dy =dy
dp
pdy
dx
+ 22 )1( 2
dydp
pp + 22 )1( 21 + 22 )1( 2ppdpdy 22 )1( 2 ppdpdy +
dydp
pp + 22 )1( 21
c
p
y +=)1(
12
