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Simple And Esy Notes On Maths And IC Engine

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Published in: Mathematics
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INFINITE SERIES,LINEAR ALGEBRA,Linear Differential Equations,ORDINARY DIFFERENTIAL EQUATIONS OF FIRST ORDER,COMPLEX NUMBERS AND ELEMENTARY FUNCTIONS OF COMPLEX VARIABLE,LAPLACE TRANSFORM,IC ENGINE MECHANISM,MULTI POINT FUEL INJECTION,PISTON INFORMATION,Testing of IC Engines

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  1. Linear Differential Equations A differential equation is linear in a set of one or more of its dependent variables if and only if each terms of the equation which contains a variable of the set or any of their derivatives is of the first degree in those variables and their derivatives A differential equation which is not linear in some dependent variable is said to be nonlinear in that variable. A differential equation which is not linear in the set of all of its dependent variables is simply said to be nonlinear. For example Differential Equation (1) (2) (3) (4) azu av ax at axz = cosx COS X — sin u — sinu Linearity is linear, ordinary and order is 2 is nonlinear sinu is linear in v but nonlinear in u . The equation is nonlinear is linear in each of the dependent variables x and y. But it is nonlinear in the set of {x, y}. The equation is nonlinear Def: A Linear differential equation is that in which the dependent variable and its derivative occur only in the first degree and are not multiplied together. Thus,the general linear differential equation of the nth order is of the form Y n + Ply n I +P2 Y pnand X are functions of x only. + pny -X ,Where PI, ,............, A Linear differential equation with constant coefficients is of the form .+ y -X ,Where PI, ,.. . pn are constant and X is either a constant or a functions of x only. dy The operator D : The part — of the symbol — may be regarded as an operator such that when dx dx it operates on y ,the result is the derivative of y. Note : If y= u is the complete solution of the equation f(D)y=0and y=v is particular solution of the equation f(D)y =x,then the complete solution of the equation f(D)y=X is y=u+v. The part y=u is called the complementary function(C.F.) and the part y=v is called the particular integral So Complete solution is y =C.F.+P.I.
  2. which is homogeneous nth-order differential equation with constant coefficients. If Y ' are m particular solutions of the above homogeneous equation, then for all real values of the parameters ' is also a solution Y = +czY2 = (3) The Characteristic Equation . Equations of the order higher than one, in general, are not solvable in terms of known functions. However, there are a number of important linear differential equations which can be solved by elementary method, and higher order ODEs with constant coefficients are one of such y = ear a family. The equation has a general form 1 y f + aoy = f (x) (1) ) the equation is nonhomogeneous nth-order differential whenf(x) is non-trivial (i.e. equation with constant coefficients. Otherwise, if the equation becomes +ax_1Y +aoy= 0 (2) To solve the homogeneous equation Eq.(16.2), substitute equation (4) which leads to a characteristic whose roots (real or complex) are . The equation Eq.( 4) is known as the characteristic equation of the differential equation Eq.( 1) or Eq.( 2). Auxilary Equation(A.E.) : The equation obtained by equating to zero the symbolic coefficient of y is called the auxilary equation. Rules for finding the Complementary function : Case(l) : if all the roots of A.E. are real and distinct .Let roots are ml, 7712, . n Then C.F. = -I-c2em2x + -BC emnx
  3. Then C.F. = e ax [(CIX + C2)cosßx +(c3x +c5em5X+ Example : Solve the following differential equations: Case(2) : if two roots of the A.E. are equal,let ml Then C.F. +%em3X+ -I-C e mnx Case(3) : if two roots of A.E.are imaginary ,let ml = u + iß and = u - iß The roots of the above equation are 1, -1 (repeated roots); hence the solution is (r Then C.F. = eaX(qcosßx +c»inßx) -kc3e17L3X A-C e mnx Case(4) : if two pair of imaginary roots be equal ,let ml = = u + iß and u - iß -kc e mnx 3 4 4 dx dx 13-61' +11L-6=O Solution : (a) The characteristic equation is The roots of the above equation are 1, 2, 3; hence the solution is y = Cle K -kc,ezx +c-se 4 (b) The characteristic equation is y = (Cl -I-czx)ex + (c-s +c4x)e- Example : Solve the following differential equation. y 5 -15/4) +84/3) - 220)?" +275y' -125)) Solution: The characteristic equation is -194 +84,3 -220r2 + 275r-125 2 2 5)
  4. In this case we've got one real distinct root, r = I , and double root, r — 5 , and a pair of complex roots, r = 2 ± i that only occur once. test +C4e2t cos(t)+ cse2t sin (t) The general solution is then, y (t) —Cl e -k C2e + C Rules for finding the particular Integral : sin(ax+b) = —sin(ax+b) Consider the differential equation (D n + all) n 1 + It can be written as f(D)y=x So PI Case(l) : When x=eax 1 e ax = —he ax , Provided f(a) 0 Then Case(2) : When X=sin(ax+b) or cos(ax+b) +an)y =x 1 Then 1 F(D2) F(a2) 1 Similarly cos (ax+b) = —cos (ax+b) F(D2) F(a2) Case(3) : When X= xm, m being a positive integer Here PI f(D) Take out the lowest degree term from f(D) to make the first term unity. The remaining factor will be of the form 1+g(D) or 1- g(D).Then take this factor in numerator.it takes the form [1 + OR [1 — g(D)]-l.Expand it in ascending powers of D as far as the term m+l m containing DTn ,since D (x ) 0 and so on.. Operate on x m term by term. Case(4) : When X =eaxv, Where V is a function of x. 1 —(eaxv) e Then
  5. Example : Solve (D — 2)2y = 8(e2X+ sin2x) Solution: A.E. is (D - -O SO D = C.F. +C2 x)e2x 1 [8(e2X+sin2x)] = 8[—!— 1 sin2x ] 1 e2x ,Put D=2•, Now 1 2(D-2) 21 Case of failure [again put D=2.case of failure ] 2 1 Also sin2x= (D —2)2 1 — Sin2x (by putting D2 = -22) - f Sin2x dx = 1 cos2x 8 4 Hence complete solution = C.F.+ P.I. The Method of Undetermined Coefficients: For each real root of multiplicity k determines k linearly independent solutions k 2 1 Since the constant coefficient i are real, complex roots occurs in , for a+ib a—ib conjugate pair, i.e. if is the root, is also the root of Eq.(4). Each complex root of multiplicity k thus yields 2k linearly independent real solutions x k-leLV cosbx eu• cosbx cosbx x k' leu• sinbx sinbx sinbx The general solution is then a linear combination of the solutions obtained above. To solve the nonhomogeneous equation Eq.( 1), (i.e. ) we ought to be able to
  6. Find one particular solution of the equation g(x). Then the complete solution is the sum of the complementary function G(x) obtained by putting in equation Eq.(l ) and g(x) i.e. (5) If f(x) is a function for which repeated differentiation yields only a finite number of linearly independent functions, appearing possibly in linear combination, then a particular integral Y for the nonhomogeneous linear equation + aoy = f (x) can be found through the procedures: (1) Assume Y to be an arbitrary linear combination of all the linearly independent functions which arise fromf(x) by repeated differentiation. (2) Substitute Y into the given differential equation. (3) Determine the arbitrary constants in Y. The class of functionf(x) possessing only a finite number of linearly independent derivatives consists of the simple functions: k (n is a positive integer) e sin cosk.x Example : Use the method of undetermined coefficients to find the complete solution of the following differential equations: 3K dx — 2y= sin2x (a) (b) dx
  7. Solution : (a) The characteristic equation is The roots of the above equation are -1, 2; hence the general solution is Y = Cle-K -I-Cze To find the particular solution, we assume that Thus, and =9Ae3X 1 . Substituting these results into the differential equation, we have 9Ae3x - 3Ae3x - 2Ae3x It follows that 1 4 1 4 that 4 and the complete solution is (b) From the results of (a), we have the general solution Y = Cle-x -ecze To find the particular solution, we assume that Yp = Asin2x+Bcos2x y = 2Acos2x —2B Y" = —4 Asin2x—4B cos2x Thus, and . Substituting these results into the differential equation, we have (—4A sin 2 x— 4B cos 2m) — (2 Ac.os 2m — 2B sin 27) — 2(A sin 2 x + B cos2x) = sin 2m or equivalently, (—6A + 2B) sin 27 + (—6B — 2 A) cos2x = (1) sin 2n- + (0) cos2x Equating coefficients of like terms, we obtain
  8. 3 20 and B = 1 20 Then the complete Solving the above system, we find that solution is 3 Y = Cle - x +cze'x• — —sin 27 +—cos2x 20 20 Example : Solve the following differential equation. y(3) - 12))" +48y' -64y = 12- 32e 8t + 2e4t Solution: We first need the complimentary solution so the characteristic equation is, 3 -12r2 +48r -64 = 4) r = 4 (multiplicity 3) We've got a single root of multiplicity 3 so the complimentary solution is, y, (t) = Cid t -kC2te4t -I-c-* e4tyc (t) = qe4t + C2te4 + c/ e Now, our first guess for a particular solution is, Notice that the last term in our guess is in the complimentary solution so we'll need to add one at least one t to the third term in our guess. Also notice that multiplying the third term by either t or t 2 will result in a new term that is still in the complimentary solution and so we'll need to multiply the third term by t 3 in order to get a term that is not contained in the complimentary solution. Our final guess is then, Y p = A + ct3e4t Now all we need to do is take three derivatives of this, plug this into the differential equation and simplify to get -64A -1728Be 8t + 6Ce4t =12-32e 8t +2e4t Setting coefficients equal and solving gives,
  9. -64A=12 -1728B = -32 6C=2 e A particular solution is then, 3 16+3 3 1 1 The general solution to this differential equation is then, 2 3 16 54 3 The Method of Variation of Parameters: In general, if in Eq.(l ) is not one of the types of functions considered above, or if the differential equation does not have constant coefficients, then we cannot apply the method of undetermined coefficients and the variation of parameters is preferred. Variation of parameters is another method for finding a particular solution of the nth- order differential equation. ax + aÆ_1 + • •A-al -I-ao (x) y = f (x) Once the solution of the associated homogeneous equation ax + + • •+al -kao (x) y = 0 (6) (7) is known. If Y ' are m particular solutions of the above homogeneous equation, then for all real values of the parameters ' is also a general solution A particular solution of Eq.( 6) has the form where y (x) , are given in Eq.(7 ) and functions of x which still must be determined. (9) , are unknown
  10. Tofind for , first solve the following linear equations simultaneously (10) Then integrate each i to obtain Vi, disregarding all constants of integration. This is permissible because we are seeking only one particular solution. The above mentioned method is known as the method of variation of parameters. Proof We are now going to prove that Eq.(10)will give us a particular solution of the differential equation Eq.( 6) for second order. When n = 2 , assume a particular solution of the form and Eq.(.8) is reduced into (x) -k 1.4 (x)yz (x) — (x) -I-vg (x)y; (x) = Then and y" y; and y (x) Substituting the above values of back to the Eq.( 1), we obtain
  11. -Bal y; +a = + ve,yz) +aoYl)+v, +aoy,) + Note : method of variation of parameters is used to find particular integral. Let P.I. = blY1 +VY2 , where u and v are unknown functions of x -f Y--u-dx and v We can use direct result u = where W is called Wronskian of Yl , y 2 And is obtained by W = Example: Use the method of variation of parameters to find the complete solution of the differential equations y" y secX Solution : symbolic form of this equation is ( +1)y = secX A.E. IS D2 +1 -O So ± C.F. = Cl cosx+ C2 sinx = Wronskian W = P.l. CIYI +C2 Y2,Where Yl =cosx = sinx and X= secx cosx —sinx sinx secx sinx cosx cosx secx dx + sinx = -cosx = -cosxf tanxdx + sinx f 1 dx =cosx logcosx + x sinx Now Complete solution is y= Cl cosx+ C2 sinx+ cosx logcosx + x sinx
  12. Cauchy's Homogeneous Linear Equation : n—l n—l An Equation Of the form xn y n -I-al X y -k. .+any -X Where a/s are constants and X is a function of x,is called Cauchy's Homogeneous linear equation. Such equation can be reduced to linear differential equation with constant coefficients by the substitution x= ez i.e., z= logx In such equations we use following operators x Y' Dy x2y" Example : Solve x2 y" 4xy' +2y 0 and so on.. Solution : Given equation is a Cauchy's homogeneous linear equation. Put x=ez i.e., z =logx And xy' Dy, x2 y" D(D-1)y , We get or (D2 + 3Dy + SO A.E. is D2 + 3Dy+ 2 Hence D = -1,-2 —2z Solution is y = Cle-z + C2e 1 1 Which is required solution of given Cauchy' s homogeneous linear equation. Legendre's Linear Equation : n—l n—l An Equation of the form (a + bx)n y n +al (a + bx) y +. .+any =X Where ai 's are constants and X is a function of x,is called Legendre's linear equation. Such equation can be reduced to linear differential equation with constant coefficients by the substitution a+bx = ez i.e., z= log(a+bx) In such equations we use following operators (a+bx) bDy
  13. (a + bX)2y" Example : Solve (1 + y" + (1 + y' +y and so on.. sin[210g(l+x)] Solution :Given equation is Legendre's linear equation. Put 1+x So z=log(l+x) And (l+x)y' Dy , (1 + x)2y" so = Sin2z Or (D2+1)y = Sin2z Which is a linear differential equation with constant coefficients. A.E. is D2+1 -O So D= C.F. = Cl cosz + c2 sinz Now P.l. Put so PI C.S. is Put z =log(l+x) Hence 1 Sin2z 1 . . —- - Sin2z 3 Cl cosz + c2 sinz - - Sin2z 1 3 Cl cos[log(l+x)]+ c2 sin[log(l+x)] 1 3 Simultaneous Linear Equations With Constant Coefficients : dx Example : Solve dt dy — 4-2k + 5 y dt
  14. Solution : — ,the given equation become (D+4)x +3y Writing D for dt And To eliminate y, operating on both sides of (1) by (D+5) and on both sides of (2) by3 and subtracting, we get Or It's A.E. is P.l. = D+5) -6]x = (D+5)t - 3et (D 2 +91) 4-14)x = 1+5t - 3et 4-91) +14 = O So D and D = -7 C.F. = Cle 2t +C2 e 1 (1+5t - 3et) D 2 +9K) +14 1 1 t-3 + D 2 4-9D +14 D 2 +91) +14 31 Hence X = Cle 2t+C2e +—t-— Similarly we can find y = 14 196 2 - Cle 2t+C2e 3 1 D 2 4-9D +14 8 9 5 1 98 24 7

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