Download A text book of engineering mathematics Volume 2 by Rajesh Pandey PDF

By Rajesh Pandey

Show description

Read or Download A text book of engineering mathematics Volume 2 PDF

Best mining books

Deepwater Petroleum Exploration & Production: A Nontechnical Guide

Textual content overviews the enterprise, engineering, and expertise of deepwater petroleum exploration and construction. presents assurance of all elements of deepwater operations: together with historical historical past; drilling and finishing wells; improvement structures; mounted buildings; floating creation structures; subsea structures; topsides; and pipelines, flowlines, and risers

Modern American Coal Mining: Methods and Applications

Glossy American Coal Mining: equipment and purposes covers an entire variety of coal mining and coal issues, with chapters written by way of prime coal mining pros and academicians. Highlights from the booklet contain coal assets and distribution, mine layout, advances in strata regulate and gear platforms, advancements in floor mining, air flow to minimize fires and explosions, drilling and blasting, staffing requirement ratios, administration and preplanning, and coal guidance and reclamation.

Essentials of Modern Open-Hole Log Interpretation

This booklet offers smooth log interpretation easily and concisely for the geologist, petrophysicist, reservoir engineer, and creation engineer accustomed to rock houses yet green with logs. It is helping you specify stable logging courses with updated instruments and interpret zones of curiosity with the most recent options.

Additional resources for A text book of engineering mathematics Volume 2

Sample text

1995) Solution. The given equation can be written as 1 dy 1 1 1 - - + - - = - log X y2 dx x Y x . 1 putting - - = v y or (1) ~ dy = dv in (1), we get y2 dx dx dv 1 1 - - - v= -logx dx x x (2) which is in the standand form of the linear equation and integrating factor dx =e- 1og x 1 e x =- -p x Multiplying both sides of (2) by the integrating factor and integrating, we get v. or ! Y 1 = (1 + log x) Y - C xy Example 15. S. 1996) Solution. Dividing both sides of the given equation by y, we get x dy + (log y) = x eX y dx - - or 1 dy 1 X - - + - (log y) = e y dx x putting v = log Y or dv 1 dy .

Mn are distinct roots of the auxiliary equation, then the general solution of (1) is Y = C I e ffijX + C 2 e ffi2X + ................ + C n em"x where CI, C2, ............ Cn are arbitrary constants Illustration. Solve the differential equation d 2y dy - 2 +3 - -54y=O dx dx Solution. The given equation is (D2 + 3D - 54) Y = 0 Here auxiliary equation is m 2 +3m-54= 0 or (m + 9) (m - 6) = 0 => m = 6,-9 Hence the general solution of the given differential equation is y = Cle6x + C2 e-9x Case II. When Auxiliary Equation has real and some equal roots.

1999) (a) Circles (b) Ellipses (c) Cycloids (d) Rectangular hyperbolas Ans. (d) 2. 1999) (a) x2y2 + 2X2 + 2y2 = C (b) x2y2 + x2 + y2 = C (c) x2y2 + X + Y = C (d) x2y2 + 2x + 2y = C Ans. (d) 3. CS. 1999) (a) 2x (y')2 + 1 = 2yy' (b) 2xy + 1 = 2yy' (c) 2x2y' + 1 = 2yy' (d) 2 (y')2 + x = 2yy' Ans. (a) 4. 2oo0) (a) xy = C eX-Y (b) x+y=Ce xy (c) xy = C eY-X (d) x-y=Ce xy Ans. (c) 34 Diiferential Equations of First Order and First 5. 2000) (a) 1 + x2 (b) (c) log (1 + X2) (d) -log (1 + x2) Ans. (a) 6. 2oo0) (a) (e6 + 9)/2 (b) (c) log e6 (d) Ans.

Download PDF sample

Rated 4.49 of 5 – based on 31 votes