General Contents
Detailed Contents
Index
Homogeneous Linear Second Order Differential Equations, Boundary Values: Example 1
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Solve
,
subject to the boundary conditions
y
(1) = 2e
-3
and
y
' (0) = -10.
General Contents
Detailed Contents
Index
See the steps of the
solution
.
The result is
y
= C
1
e
-3
x
+ C
2
*
x
*
e
-3x
.
What do we mean by “boundary conditions”?
We mean a specification of the value of
y
and
y
' at certain values of the independent variable.
In many problems, the independent variable is position, and the boundary conditions refer to the boundary coordinates of a system.
How do boundary conditions help?
They lead to equations for determining C
1
and C
2
.
How do we make use of
y
(1) = -2e
-3
?
We substitute
x
= 1 into our equation for y to get an equation involving C
1
and C
2
.
Do it.
Simplify
y
(1).
y
(1) =e
-3
* [C
1
+ C
2
] = –2 * e
-3
, or
C
1
+ C
2
= -2
This is one of the equations we can use to find the values of these coefficients.
To get another equation, we can use the other boundary condition.
How do we use
y
' (0) = –10?
We need to differentiate
y
to get
y
', and then substitute x = 0.
Use the Product Rule and Chain
Rule to find
y
'(
x
).
Do the indicated derivatives.
Determine
y
' (0).
So we have two equations from these two boundary conditions:
C
1
+ C
2
= –2
-3C
1
+ C
2
= –10
How do we solve for the C’s?
One way is to subtract the second equation from the first.
Do that.
C
1
+ C
2
= –2
-3C
1
+ C
2
= –10
4C
1
= 8, or
C
1
= 2.
Substitute this in C
1
+ C
2
= –2.
We get 2 + C
2
= –2, or
C
2
= – 4.
Use these values for C
1
and C
2
to get a solution for y.
We use them in our general solution above and get
y(x)
= C
1
e
-3x
+ C
2
e
-3x
=
2e
-3x
– 4
x
*
e
-3x
As a check, confirm that
y
(1) = –2e
-3
.
y
(1) =
2e
-3
– 4
e
-3
= –2e
-3
, which checks.
Check that
y
'(0) = –10
From
we get
y
'(0) = e
0
* {-6 - 4*[0 + 1]} =
–10, which checks.
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