HP 50g User Manual Page 483

Graphing calculator.

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
K
e
+ (450 x
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
p
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' `
Allow the calculator about ten seconds to produce the result: 'X^2 = X^2'.
Example 3 - Solving a system of linear differential equations with constant
coefficients.
Consider the system of linear differential equations:
2
+330 x+241)/13500.
h
y(x) = y
2
= (450 x
+330 x+241)/13500, is indeed a particular solution
SUBST
x
1
represents the solution to the
h
–3x
= K
e
+ K
1
2
(x), the solution of the
h
(x) + y
(x),
h
p
EV L
'(t) + 2x
'(t) = 0,
2
e
1
5x
2x
e
+ K
e
. You can prove
3
=
p
–3x
5x
+ K
e
+
2
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