HP 48gII User Manual Page 484

Graphing calculator.

of constants result from factoring out the exponential terms after the Laplace
transform solution is obtained.
Example 2 – Using the function LDEC, solve the non-homogeneous ODE:
3
d
y/dx
Enter:
'X^2' ` 'X^3-4*X^2-11*X+30' ` LDEC
The solution, shown partially here in the Equation Writer, is:
Replacing the combination of constants accompanying the exponential terms
with simpler values, the expression can be simplified to y = K
2x
2
K
⋅e
+ (450⋅x
+330⋅x+241)/13500.
3
We recognize the first three terms as the general solution of the homogeneous
equation (see Example 1, above). If y
homogeneous equation, i.e., y
that the remaining terms in the solution shown above, i.e., y
2
(450⋅x
+330⋅x+241)/13500, constitute a particular solution of the ODE.
Note: This result is general for all non-homogeneous linear ODEs, i.e., given
the solution of the homogeneous equation, y
corresponding non-homogeneous equation, y(x), can be written as
where y
(x) is a particular solution to the ODE.
p
To verify that y
= (450⋅x
p
solution of the ODE, use the following:
'd1d1d1Y(X)-4*d1d1Y(X)-11*d1Y(X)+30*Y(X) = X^2'`
'Y(X)=(450*X^2+330*X+241)/13500' `
3
2
2
-4⋅(d
y/dx
)-11⋅(dy/dx)+30⋅y = x
represents the solution to the
h
–3x
5x
= K
⋅e
+ K
⋅e
h
1
2
(x), the solution of the
h
y(x) = y
(x) + y
(x),
h
p
2
+330⋅x+241)/13500, is indeed a particular
SUBST
EVAL
2
.
–3x
5x
⋅e
+ K
⋅e
1
2
2x
+ K
⋅e
. You can prove
3
=
p
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