
y" = .5j/' + ,5y + M+1
Step 1: Convert the equation into a first-order equation:
Example: Solve this equation for tti(l) given that y{Q) = 0 and
l/'(0) = 0 («;(0) = [ 0 0 ]):
y
0 1
.5 .5
y
(.51+1)
Step 2.
Store the values in fw (
0 1
.5 .5
and c (
OUDSKID 0 dED 1
(B -5 CIED -5 iEFrlRl CD ® © FW (3 (STO)
SICED
0
fSPC)
1
(ENTER
1 CD
©
c
(STO"|
.jg Step 3: Enter the equation and initial values, set the solution
variable to w, and solve for w{\:y.
f?»)fSOLVr| |T| UK
(3 (3 FW ® CD F' © ®
(30 .5 © © T C£) 1
ENTER
SOLVE
F- 'FW*W+C*<.5*T+D'
INOEP: T INIT: 0 FINftL: 1
SDLN: Id INIT: [ 0,„ FINULtP
TDL:.0001 STEP: Df It _ STIFF
PRESS SOLVE FDR FINAL SOLN VALUE
H^aiiiHiiiiiiiiMiiiiiiBaBiiiNtaRi^^
© T (ENTER)0(ENTER) 1
(ENTER) © W (ENTER) QCLD
0 fSPCT) 0 (ENTER) S0LVE
Press to view the result vector, w;(l), [ ,718262064225
1.71826206422 ]. The first value is ?/(l), the second value is :?/'(l).
How accm-ate is the answer? The original equations are
y = e^-t-l
and
y' = t + y
Evaluating the equations at 1 and comparing the results you can see
there is an error of approximately 0.0000198, which is well within the
specified error tolerance of 0.0001.
19-6 Differential Equations
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