General Contents
Detailed Contents
Index
Homogeneous Linear Second Order Differential Equations, Initial Values: Example 2
If you find this page helpful and would recommend that I create more pages like this one, please let me know:
Email to John Taylor
If you want to see all of the following steps at once, click the "All Steps" button. Otherwise, use the "Next" button.
Solve
,
subject to the initial conditions
y
(0) = – 4 and
y
' (0) = 8.
General Contents
Detailed Contents
Index
See the steps of the
solution
.
The result is
y
= C
1
e
3x
cos(
4x
) + C
2
e
3x
sin(
4x
).
What do we mean by “initial conditions”?
We mean a specification of the value of
y
and
y
' at the origin, or when the independent variable is 0.
In many problems, the independent variable is time, which we think of starting at 0 and increasing.
How do initial conditions help?
They lead to equations for determining C
1
and C
2
.
How do we make use of
y
(0) = – 4?
We substitute
x
= 1 into our equation for y to get an equation involving C
1
and C
2
.
Do it.
What are the values of
e
0
, cos(0) and sin(0)?
Any base to the power 0 equals 1; cos(0) = 1; sin(0) = 0.
Simplify
y
(0).
y
(0) = C
1
*1*1 + C
2
*0 = – 4 = C1.
This is one of the equations we can use to find the values of the coefficients.
To get another equation, we can use the other initial condition.
How do we use
y
' (0) = 8?
We need to differentiate
y
to get
y
', and then substitute x = 0.
Use the Product Rule to find
y
'(
x
).
Use the Chain Rule to do the indicated derivatives.
Determine
y
' (0).
So we have two equations from these two initial conditions:
C
1
= – 4
3C
1
+ 4C
2
= 8
How do we solve for the C’s?
We can substitute C1 = – 4 into the second equation.
Do that.
3*(– 4) + 4C
2
= 8, or
C
2
= 5
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
= C
1
e
3x
cos(4
x
) + C
2
e
3x
sin(4
x
)
=
– 4e
3x
cos(4
x
) + 5
e
3x
sin(4
x
)
As a check, confirm that
y
(0) = – 4.
y
(0)= – 4*e
0
*cos(0) + 5
e
0
sin(0) = – 4 + 3*(0) = – 4, which checks.
Check that
y
'(0) = –1
From
we get
y
'(0) = – 4*[ e
0
(-sin(0) * 4) + cos(0) * e
0
*3 ] + 5 * [ e
0
* cos(0) * 4 + sin(0) * e
0
* 3 ], or
y
'(0) = – 4 * 3 + 5 * 4 = 8, which checks.
The end. If you found this helpful and would recommend that I create more pages like this one, please let me know:
Email to John Taylor
,