Consider the initial value problem e9 tey sin ydydt 0 y2

Consider the initial value problem e^9 + (te^y - sin y)dy/dt = 0, y(2) = 1.5. Use ode4 5 to find approximate values of the solution at x = 1, 1.5, and 3. Then plot the solution on the interval 0.5 lessthanorequalto t lessthanorequalto 4. In this pan you should use the results from parts (c) and (d) of Problem 7 in Problem Set B. Compare the values of the actual solution and the numerical solution at the three specified points. Plot the actual solution and the numerical solution on the same graph. Now plot the numerical solution on several large intervals (e.g., 2 lessthanorequalto t lessthanorequalto 10, or 2 lessthanorequalto t lessthanorequalto 100, or even 2 lessthanorequalto t lessthanorequalto 1000). Make a guess about the nature of the solution as t rightarrow infinity. Try to justify your guess on the basis of the differential equation. Similarly, plot the numerical solution on intervals t lessthanorequalto t lessthanorequalto 2 for different choices of as a small positive number. What do you think is happening to the solution as t rightarrow 0+?

Solution

.MODEL SMALL .STACK 100H .DATA PROMPT DB \'Current System Time is : $\' TIME DB \'00:00:00$\' ; time format hr:min:sec .CODE MAIN PROC MOV AX, @DATA ; initialize DS MOV DS, AX LEA BX, TIME ; BX=offset address of string TIME CALL GET_TIME ; call the procedure GET_TIME LEA DX, PROMPT ; DX=offset address of string PROMPT MOV AH, 09H ; print the string PROMPT INT 21H LEA DX, TIME ; DX=offset address of string TIME MOV AH, 09H ; print the string TIME INT 21H MOV AH, 4CH ; return control to DOS INT 21H MAIN ENDP ;**************************************************************************; ;**************************************************************************; ;------------------------- Procedure Definitions ------------------------; ;**************************************************************************; ;**************************************************************************; ;**************************************************************************; ;------------------------------ GET_TIME --------------------------------; ;**************************************************************************; GET_TIME PROC ; this procedure will get the current system time ; input : BX=offset address of the string TIME ; output : BX=current time PUSH AX ; PUSH AX onto the STACK PUSH CX ; PUSH CX onto the STACK MOV AH, 2CH ; get the current system time INT 21H MOV AL, CH ; set AL=CH , CH=hours CALL CONVERT ; call the procedure CONVERT MOV [BX], AX ; set [BX]=hr , [BX] is pointing to hr ; in the string TIME MOV AL, CL ; set AL=CL , CL=minutes CALL CONVERT ; call the procedure CONVERT MOV [BX+3], AX ; set [BX+3]=min , [BX] is pointing to min ; in the string TIME MOV AL, DH ; set AL=DH , DH=seconds CALL CONVERT ; call the procedure CONVERT MOV [BX+6], AX ; set [BX+6]=min , [BX] is pointing to sec ; in the string TIME POP CX ; POP a value from STACK into CX POP AX ; POP a value from STACK into AX RET ; return control to the calling procedure GET_TIME ENDP ; end of procedure GET_TIME ;**************************************************************************; ;------------------------------- CONVERT --------------------------------; ;**************************************************************************; CONVERT PROC ; this procedure will convert the given binary code into ASCII code ; input : AL=binary code ; output : AX=ASCII code PUSH DX ; PUSH DX onto the STACK MOV AH, 0 ; set AH=0 MOV DL, 10 ; set DL=10 DIV DL ; set AX=AX/DL OR AX, 3030H ; convert the binary code in AX into ASCII POP DX ; POP a value from STACK into DX RET ; return control to the calling procedure CONVERT ENDP ; end of procedure CONVERT