HP 50G

Page 16-19
With Y(s) = L{y(t)}, and L{d
2
y/dt
2
} = s
2
⋅Y(s) - s⋅y
o
– y
1
, where y
o
= h(0) and y
1
= h’(0), the transformed equation is
s
2
⋅Y(s) – s⋅y
o
– y
1
+ 2⋅Y(s) = 3/(s
2
+9).
Use the calculator to solve for Y(s), by writing:
‘X^2*Y-X*y0-y1+2*Y=3/(X^2+9)’ ` ‘Y’ ISOL
The result is
‘Y=((X^2+9)*y1+(y0*X^3+9*y0*X+3))/(X^4+11*X^2+18)’.
To find the solution to the ODE, y(t), we need to use the inverse Laplace
transform, as follows:
OBJ ƒ ƒ Isolates right-hand side of last expression
ILAPμ Obtains the inverse Laplace transform
The result is
i.e.,
y(t) = -(1/7) sin 3x + y
o
cos √2x + (√2 (7y
1
+3)/14) sin √2x.
Check what the solution to the ODE would be if you use the function LDEC:
‘SIN(3*X)’ ` ‘X^2+2’ ` LDEC μ
The result is:
i.e., the same as before with cC0 = y0 and cC1 = y1.
Note: ‘SIN(3*X)’ ` LAP μ produces ‘3/(X^2+9)’, i.e.,
L{sin 3t}=3/(s
2
+9).