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Let satisfy the differential equation for with initial values and Find the exact values of and A Bernoulli Equation Solve the IVP with Show that the general solution of is Show that the constant C...

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Let satisfy the differential equation for with initial values and Find the exact values of and
A Bernoulli Equation
Solve the IVP with
Show that the general solution of is Show that the constant C must be positive, and that if then the solution will have vertical asymptotes and What will the domain of such solutions look like? Sketch their graphs for and for
Explain how the derivation of the general solution fails to obtain the obvious solution for all x. In fact there is no value of x for which How can this be?
Solve the IVP with [Hint: Find an IVP for the new variable
Discuss the IVP with In what way are the solutions of this problem different from those of 2(b)?
Write for the linear differential operator
Verify that is a solution of the equation for all
Use the fact that to find a linearly independent function which is also a solution of the equation for all
Use the fact that and to write the general
solution of for all
Now solve the initial value problem for all
Solve the following first order differential equations:
(Find the general solution.)
with
(Find the general solution.)
A Second Order Equation with Constant Coefficients
Write the differential equation as and complete the square to find constants a and b such that
Find a constant c and such that
Use (b) to find the general solution of
Find any particular solution to
Use (c) and (d) to find the general solution to
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Let satisfy the differential equation for with initial values and Find the exact values of and A Bernoulli Equation Solve the IVP with Show that the general solution of is Show that the constant C must be positive, and that if then the solution will have vertical asymptotes and What will the domain of such solutions look like? Sketch their graphs for and for Explain how the derivation of the general solution fails to obtain the obvious solution for all x. In fact there is no value of x for which How can this be? Solve the IVP with [Hint: Find an IVP for the new variable Discuss the IVP with In what way are the solutions of this problem different from those of 2(b)? Write for the linear differential operator Verify that is a solution of the equation for all Use the fact that to find a linearly independent function which is also a solution of the equation for all Use the fact that and to write the general solution of for all Now solve the initial value problem for all Solve the following first order differential equations: (Find the general solution.) with (Find the general solution.) A Second Order Equation with Constant Coefficients Write the differential equation as and complete the square to find constants a and b such that Find a constant c and such that Use (b) to find the general solution of Find any particular solution to Use (c) and (d) to find the general solution to

005_sgzjpk-54lldotv.doc
Answered Same Day Dec 21, 2021

Solution

David answered on Dec 21 2021
129 Votes
Question: 3 (i):
Solution 1: Given the differential equation
0
=
+
¢
¢
y
y
x
for
,
1
1
-
x
with initial values
(
)
0
1
=
y
and
(
)
.
1
1
=
¢
y
Let
(
)
Ã¥
Â¥
=
-
=
0
1
n
n
n
x
a
y
.
Taking the first derivative of y as y’’.
(
)
1
1
'1
n
n
n
ynax
Â¥
-
=
=-
Ã¥
Hence, the second derivative, y’’ will be:
(
)
2
2
''(1)1
n
n
n
ynnax
Â¥
-
=
=--
Ã¥
Plugging these two into the differential equation.
(
)
(
)
2
20
(1)110
nn
nn
nn
xnnaxax
¥¥
-
==
--+-=
åå
The first part of this equation has to be converted from an order two to zero.
(
)
(
)
2
00
(1)(2)110
nn
nn
nn
xnnaxax
¥¥
+
==
++-+-=
åå
(
)
2
0
(2)(1)(1)0
n
n
nn
n
xnnaax
Â¥
+
=
+++-=
Ã¥
Here,
2
(2)(1)
nn
nnaa
+
+++
has to be equal to zero where, n = 0, 1, 2…
Solving, we get:
2
(2)(1)
nn
xnnaa
+
++=-
Or,
2
(2)(1)
n
n
a
a
xnn
+
-
=
++
We know,
,
1
1
-
x
So, x = 2
Now putting values for obtaining values of a1, a2, a3, a4 etc
2
(2)(1)
n
n
a
a
xnn
+
-
=
++
For n = 0,

0
2
2(2)(1)
a
a
-
=
For n = 1,
11
3
2(4)(3)12
aa
a
--
==-
For n = 2,
2
4
24
a
a
-
=
, etc.
Solution 2: (a): Solve
0
5
=
+
+
¢
xy
xy
y
for
(
)
.
1
1
=
y
This can be solved to give:
(b) General solution of
0
5
=
+
+
¢
xy
xy
y
from the above calculation is
(
)
.
1
4
1
2
2
-
-
=
x
Ce
y
For
(
)
.
1
1
=
y
2
11,
2*7.38
,0.2701
Ce
C
orC
=-
=
=
Given that the vertical asymptotes
(
)
C
x
1
ln
2
1
=
and
(
)
.
1
ln
2
1
C
x
-
=
Put C = 0.2701 for obtaining x for both asymptotes.
X = 0.529, -0.529
The graph can be plotted as:
(c)
This can be solved to give:
(
)
.
1
4
1
2
2
-
-
=
x
Ce
y
For
(
)
0
=
x
y
,...
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