Saturday, 11 June 2016

Assignment of differential equation It should be neat embedded in file cover Assignment folder Otherwise 0 Marks

Q#1:

x^3-3x^2y=c 
3x^2 dx-3(2xydx+x^2dy)=0
3x^2 dx-6xy dx-3x^2 dy=0
 Divide by 3x
xdx-2ydx-xdy=0
xdx-2ydx-xdy=0
(x-2y)dx-xdy=0



Q#3:
Solution for (x^5+3y)dx-xdy=0 equation:

Simplifying
(x5 + 3y) * dx + -1xdy = 0

Reorder the terms for easier multiplication:
dx(x5 + 3y) + -1xdy = 0
(x5 * dx + 3y * dx) + -1xdy = 0

Reorder the terms:
(3dxy + dx6) + -1xdy = 0
(3dxy + dx6) + -1xdy = 0

Reorder the terms:
3dxy + -1dxy + dx6 = 0

Combine like terms: 3dxy + -1dxy = 2dxy
2dxy + dx6 = 0

Solving
2dxy + dx6 = 0

Solving for variable 'd'.

Move all terms containing d to the left, all other terms to the right.

Factor out the Greatest Common Factor (GCF), 'dx'.
dx(2y + x5) = 0



Q#4:

Solution for vdx+(2x+1-vx)dv=0 equation:

Simplifying
vdx + (2x + 1 + -1vx) * dv = 0

Reorder the terms:
dvx + (1 + -1vx + 2x) * dv = 0

Reorder the terms for easier multiplication:
dvx + dv(1 + -1vx + 2x) = 0
dvx + (1 * dv + -1vx * dv + 2x * dv) = 0

Reorder the terms:
dvx + (1dv + 2dvx + -1dv2x) = 0
dvx + (1dv + 2dvx + -1dv2x) = 0

Reorder the terms:

1dv + dvx + 2dvx + -1dv2x = 0


Combine like terms: dvx + 2dvx = 3dvx
1dv + 3dvx + -1dv2x = 0

Solving
1dv + 3dvx + -1dv2x = 0

Solving for variable 'd'.


Move all terms containing d to the left, all other terms to the right.

Factor out the Greatest Common Factor (GCF), 'dv'.
dv(1 + 3x + -1vx) = 0


Thanks and Regards
Ahad bajwa +923336773758                               Ibrahim Farid +923335068377

Course Outline of Operating System

Course outline of O.S

CHP 1 R.Q  Q1,2,5,18,15,13,22,29.33.39
CHP 2 R.Q Q4,5,6,7,15,17,19,25,28,29,37,39
.
CHP 4 R.Q Q1,2,3,4,5,9,14,16,20,29,30
CHP 6 R.W Q1,2,3,4,6,7,8,9,11,12 + BANKER'S ALGORITHM.

Monday, 6 June 2016

SQL Query Of Today's Quiz 6/6/2016

SQL Query Of Today's Quiz


create table  CUSTOMER786(
cUSTOMER_ID vARCHAR(20) NOT NULL,
nAME vARCHAR(20),
c_ADDRESS VARCHAR(20),
cONTACT INT NOT NULL,
c_ORDER_ID vARCHAR(20)
pRIMARY KEY (cUSTOMER_ID,C_oRDER_ID))
INSERT INTO CUSTOMER786 VALUES
('cs-113','SLEEM','ISB',21457,'O-114');
 SELECT * FROM CUSTOMER786;
 SELECT cUSTOMER786.nAME,cUSTOMER786.c_ADDRESS,ORDERS786.No_OF_ITEM
 FROM CUSTOMER786,oRDERS786;
 uPDATE CUSTOMER786
 sET C_ADDRESS='rWP'
 WHERE nAME='ISB';
CREATE TABLE ORDERS786(
C_ORDER_ID VARCHAR(20),
CUSTOMER_ID VARCHAR(20),
nO_OF_ITEM INT NOT NULL,
pRODUCT_ID vARCHAR(20),
pRODUCT_dESCRIPTION VARCHAR(20),
aMOUNT INT NOT NULL,
PRIMARY KEY(pRODUCT_ID),
fOREIGN KEY(cUSTOMER_ID,C_oRDER_ID) REFERENCES cUSTOMER786(cUSTOMER_ID,C_oRDER_ID));
aLTER TABLE ORDERS786
aDD MANUFACTURERS vARCHAR(20);
INSERT INTO ORDERS786 VALUES('O-112','CS-111',10,'P-113','SOAP',300,'man');
SELECT * FROM ORDERS786;
sELECT CUSTOMER_ID FROM ORDERS786
WHERE pRODUCT_dESCRIPTION='sOAP' aND pRODUCT_dESCRIPTION='SHAMPOO';
cREATE TABLE pRODUCT786(
pRODUCT_ID VARCHAR(20),
pRODUCT_dESCRIPTION vARCHAR(20)
pRIMARY KEY (pRODUCT_ID));
INSERT INTO pRODUCT786 VALUES('P-114','TOOTH PAST');
SELECT * FROM pRODUCT786;

Diff Eq Notes

Diff Eq Notes

Over 140 MB So Sit Tight

Sunday, 5 June 2016

Data Base SQL Code

Command Done in Data Base
1)SELECT column_name,column_name
FROM table_name;

AND

SELECT * FROM table_name;

2)

SQL SELECT DISTINCT Syntax

SELECT DISTINCT column_name,column_name
FROM table_name;

3)

SQL WHERE Syntax

SELECT column_name,column_name
FROM table_name
WHERE column_name operator value;

4)
AND Operator Example

SELECT * FROM Customers
WHERE Country='Germany'
AND City='Berlin';

OR Operator

SELECT * FROM Customers
WHERE City='Berlin'
OR City='München';

5)
SQL INSERT INTO Syntax
INSERT INTO table_name
VALUES (value1,value2,value3,...);

6)

SQL UPDATE Syntax

UPDATE table_name
SET column1=value1,column2=value2,...
WHERE some_column=some_value;

7)

SQL DELETE Syntax

DELETE FROM table_name
WHERE some_column=some_value;

8) TRUNCATE Table Name;

9)

SQL Alias Syntax for Columns

SELECT column_name AS alias_name
FROM table_name;


10)

SQL Alias Syntax for Tables

SELECT column_name(s)
FROM table_name AS alias_name;

11)

SQL FULL OUTER JOIN Syntax

SELECT column_name(s)
FROM table1
FULL OUTER JOIN table2
ON table1.column_name=table2.column_name;

12)

SQL PRIMARY KEY Constraint on CREATE TABLE


CREATE TABLE Persons
(
P_Id int NOT NULL,
LastName varchar(255) NOT NULL,
FirstName varchar(255),
Address varchar(255),
City varchar(255),
PRIMARY KEY (P_Id)
)

13) SQL FOREIGN KEY Constraint on CREATE TABLE

CREATE TABLE Orders
(
O_Id int NOT NULL,
OrderNo int NOT NULL,
P_Id int,
PRIMARY KEY (O_Id),
FOREIGN KEY (P_Id) REFERENCES Persons(P_Id)
)

14)The DROP TABLE Statement

DROP TABLE table_name;

15)

SQL ALTER TABLE Syntax

To add a column in a table, use the following syntax:
ALTER TABLE table_name
ADD column_name datatype;

16)

SQL CREATE VIEW Syntax

CREATE VIEW view_name AS
SELECT column_name(s)
FROM table_name
WHERE condition

Example

CREATE VIEW [Current Product List] AS
SELECT ProductID,ProductName
FROM Products
WHERE Discontinued=No;

We can query the view above as follows:
SELECT * FROM [Current Product List];

These are the 16 queries which Ma'am Taught.

Thanks And Regards

Saturday, 4 June 2016


http://www.2shared.com/file/EFSm7PYD/Ise_slides.html
SLIDES OF I.S.E
5 Questions
5 Presentation
Already concise d the slides on requesting the SIR

Every body is requested to Attend the class ON MONDAY 1.30-3.00 in which discussion Regarding paper pattern will be held.


Thanks and Regards

Ahad Bajwa ISPR                                                Ibrahim Farid
                                           
+923336773758                                                    +923335068377

Friday, 3 June 2016

SQL CODE # 2

Create Table Book(
Book_Id int not null,
Book_Name Varchar(20),
Author_Name Varchar(20),
Edition Varchar(20),
Publisher Varchar(20),
User_id Varchar(20),
Issue_date Varchar(20)
Primary Key(Book_id,User_id))

Alter table book
Add Return_date Varchar(20);
insert into Book values
(001,'Database','Ahsan Raza','2020','Abc','S0012','2016');
select * from Book;
select USER_ID from Book
where Book_Name='Operating System';

Create table user_id(
USER_Id Varchar(20) not null,
Name varchar(20),
Dept Varchar(20),
program Varchar(20)
Primary key(User_id)
Foreign key(User_id)references User_id(User_id));

insert into user_id values('s011','ali','c.s','mscs');

select * from user_id;


select book.Book_Id, user_id.USER_Id
from Book,user_id
where book.User_id=user_id.user_id;

Differential Equation Assignment # 2


Assembly Language

Slides of Assembly Language

Files can be downloaded by clicking on above mentioned hyperlink