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Simple And Esy Notes On Maths And IC Engine

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  • Prashaant

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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

  • 1
    ORDINARY DIFFERENTIAL EQUATIONS OF FIRST ORDER Def :- Differential equations are the equations containing derivatives or differentials of one or more variables, for example the followings are differential equations: æ=ex +sinx (2) y" - 2y' + y = cosx (3) (4) 2x2dx+ 2ydy = 0 An ordinary differential equation (ODE) is an differential equation in which all derivatives of one or more dependent variables with respect to a single independent variable. Clearly, equations (1), (2),and (4) are ODEs, while (3) is not. In fact, (3) is a partial differential equation. A partial differentialequation (PDE) is a differential equation containing at least one partial derivative of some dependent variable. Definition : A solution of differential equation is the known function which satisfies the equation. Definition: The order of differential equations is defined as the order of the highest-order derivative which appears in the equation. Definition: A differential equation along with subsidiary conditions on the unknown function and its derivatives, all given at the same value of the independent variable, constitutes an initial-value problem. The subsidiary conditions are initial conditions. If the subsidiary conditions are given at more than one value of the independent variable, the problem is a boundary-value problem and the conditions are boundary conditions. Exact Differential Equation : An differential equation M(x, y)dx + N(x, y)dy ax and exact differential equation if and only if there is a function f such that —dx+—dy = 0 is an throughout some region. Thus, we have or d[f(x, y)] = 0. The solution isf(x, y) = k, k is the arbitrary constant.
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    Although some exact equations are easy to identify and solve, it is general impossible to tell by inspection whether a given first-order differential equation is exact or not. Theorem : ax If and are continuous, then the differential equation M(x, y)dx + N(x, y)dy = 0 is exact if and only if Proof : If the differential equation is exact, there exists a differential functionf(x, y) for which O. we have and as a condition of exactness. If, in addition, M and N are differentiable, it follows that ayax axay awf whenever the mixed partial derivatives of f exist and are continuous. Hence, whenever and exists, are continuous, and are equal. ansf To prove the converse of the theorem, we assume that Y any Then there is a function f such that and Let us first integrate M(x, y) with respect to x, holding y fixed. Introducing the dummy variable t gives us the expression Our proof will be completed if we can determine c(y) so that Y We have from (1)
  • 3
    dt +c'(y) = at Thus Y will equal to N(x, y), as required, if c(y) is determined so that c '(y) = N(xo, y), that is, if c Integrating Factor: Sometimes, a non-exact differential equation M(x, y)dx + N(x, y)dy = 0 can be turned into an exact differential equation by multiplying the whole equation by an appropriate factor, called an integrating factor. The following observations are often helpful in finding integrating factor:- (1) If a first-order differential equation contains the combination xdx+ydy = —d(x' + y try some function x2+ y as a multiplier. (2) If a first-order differential equation contains the combination xdy + ydx = d(xy), try some function xy as a multiplier. 1 —T (3) If a first-order differential equation contains the combination xdy - ydx, try x or 1 1 as a multiplier. If neither of these works, try x + Y or some function of -1 Y — ydx d tan these expressions, as an integrating factor, remember that and xdy — ydx tan - Also, for the general first-order differential equation M(x, y)dx + N(x, y)dy = 0, the following theorem indicates other possibilities of finding integrating factor.
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    ay ax (a) If is a function of x only, say f(x), then integrating factor of the equation M(x, y)dx + N(x, y)dy = 0 1 ant any is a function of y only, say g(y), then (b) if integrating factor of the equation M(x, y)dx + N(x, y)dy = 0 Example : is an is an Show that the following differentials are exact and solve the corresponding differential equations: (a) (9x2 + y - l)dx - (4y - x)dy = 0 (b) (exsiny - 2ysinx)dx + (excosy + 2cosx)dy Solution : (a) For this ODE, The ODE is exact. The differential function is ay 3 By comparing the above equations, we have f(x, y) = 3x3+ xy - x - 2y 2 Therefore the general solution of the ODE is 3x3+ xy - x - 2y2 = k
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    (b) For this ODE, an,f M (x, y) = ex sin y — 2 y sinx = e x cosy— 2 sin x = ex cosy+2cosx = cosy—2sinx ax The ODE is exact. The differential function is = ex sin y — 2y sin x f (x, y) = ex sin y + 2y cosx+C1(y) = e K cosy+2cosx f (x, y) = ex By comparing the above equations, we have f(x, y) = exsiny + 2ycosx Therefore the general solution of the ODE is exsiny + 2ycosx = k Example : Find the integration factor of the following ODE and solve the corresponding equation (3x2y+ 2xy + y3)dx + (x2 + y2)dy = 0 Solution : For this ODE, M (x, y) = 3m 2 y + 2xy + y 3 -2} = 3x2 + 2 x+3yl and = 27,- an,f = 6xy + 2y ax and
  • 6
    We can see that depends on variable x only. Therefore the exp integration factor is given by e3X(3x2y+ 2xy + y3)dx + e3X(x2 + y2)dy The differential function isf(x, y) (37 3) 3 — = + 3e3x x 'y + Y 3 3 and the ODE is reduced into 3 3 + Cl(x) By comparing the above equations, we have f(x, y) = e3X(2xy+ 3x2y + y3) Therefore the general solution of the ODE is e3X(2xy+ 3x2y + y3) = k Separable First-Order Differential Equations: A separable differential equation is a first-order ordinary equation that is algebraically reducible to a standard differential form in which each of the non-zero terms contains exactly one variable solutions to this kind of equation is usually quite straightforward. For example f(x)dx +g(y)dy = 0, the solution will be, of course One way of discovering whether or not a given equation is separable is to collect coefficients on the two differentials and see if the result can be put in the form f(x)G(y)dx = F(x)g(y)dy Another way is to solve for a derivative and compare the result with
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    dy A general solution of the form f(x)G(y)dx = F(x)g(y)dy can be found by first dividing by the product F(x)G(y) to separate the variables and then integrating dx = dy Similarly, the second form can be solved by first multiplying by dx, then dividing by N(y) and subsequently integrating = M (x)dx +c The process of solving a separable equation will often involve division by one or more expressions. In such cases the results are valid where the divisors are not equal to zero but may or may not bemeaningful for values of the variables for which the division is undefined. Such values require special consideration and may lead to singular solutions! Example: Solve the following differential equations by separation of variables: dy (a) dy dx (b) (1 dy Solution: (a) The ODE Y becomes ydy = x2dx and
  • 8
    yd y = 7,3dx dy dx 2 3 3 3 + 2C 3 (b) The ODE becomes and 2 3 3 3 Special First-Order Equations: Bernoulli Equation: A Bernoulli differential equation has the form Clearly, for n = 0 and 1, the equation is linear; for other values of n, it is nonlinear. However, changing the dependent variable dz Example: Solve the Bernoulli equation converts it into the linear equation x Solution: 2xy— The ODE dz we have Hence 2y dx becomes dz dy d yd x 1 —y . By changing the variable
  • 9
    1 d 7,- 'z = x 2 + Cn- and Clairaut's Equation : dz 1 dx and its integration factor is Therefore, we obtain xp(J 1 1 exp z + Cm An Equation of the form y =px +f(p) is known as Clairaut' s Equation Note : By putting p = c in above, the required solution is y = cx +f(c),Thus the solution of Clairaut's equation is obtained by putting c for p. Example: Solve (y-px)(p-l) = p Solution : The given equation can be written in form y-px =p/p-l Or y =px+p/p-l Which is Clairaut's equation.hence putting c for p ,the solution is y cx+c/c-l Leibnitz's linear equation: dy The general form of linear differential equation of first order is — + py = Q ,where p and Q are dx functions of x only or may be constants. (1) if , it is solvable by direct integration, or (2) if , the equation is separable. Theorem: dy dx The equation as an integration factor.
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    xp(J Proof. When eq is multiplied by vvexp(J = which has solution , it can be written in the form So, we have the following steps to solve a linear first-order differential equation The following procedures are often helpful in finding solution to the equations:- (1) Compute the integrating factor. (2) Multiply the right-hand side of the given equation by this factor and write the left-hand side as the derivative of y times the integrating factor. (3) Integrating and solving the equation for y. Example: Solve the following differential equations by integration factor method: xu— ky x dx (a) dy — + y tan x = sec x Solution: dy a) The ODE k.dx exp dy becomes The integration factor is -k.
  • 11
    The ODE can be reduced to sec = tan x + C — + y tan x sec x = sec x ..y=sm x + Ccosx sec. 2 — ) — sec x — sec' x

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