ENGINEERING MATHEMATICS-II. [As per Choice Based Credit System (CBCS) scheme]. (Effective from the academic year ). SEMESTER - I/II. Engineering Mathematics-II. (Common to all Branches). Time: 3 Hrs. nbafinals.info: Note: Answer any FIVE full questions, choosing at least ONE question from. Download All These Question Papers in PDF Format, Check the Below Table to VTU BE Engineering Mathematics - II Question Papers.
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Engineering Class handwritten notes, exam notes, previous year questions, PDF free download. Note for Applied Mathematics - 2 - M-2 By vtu rangers. VTU Engineering Maths II 15mat21 Solution - Download as PDF File .pdf), Text File .txt) or read online. Engineering mathematics 2 sample question with. This book Engineering Mathematics-II is designed as a self-contained, comprehensive classroom text for the second semester B.E. Classes of Visveswaraiah.
Laplace Transform, Inverse Laplace Transform, Linearity, transform of derivatives and Integrals, Unit Step function, Dirac delta function, Second Shifting theorem, Differentiation and Integration of Transforms, Convolution, Integral Equation, Application to solve differential and integral equations, Systems of differential equations. Power series; radius of convergence, power series method, Frobenius method; Special functions: Vector and Scalar functions and fields, Derivatives, Gradient of a scalar field, Directional derivative, Divergence of a vector field, Curl of a vector field.
Finding the current in electrical circuits.
Eigen values — Eigen vectors — Properties — Cayley-Hamilton theorem Inverse and powers of a matrix by using Cayley-Hamilton theorem- Diagonalization- Quadratic forms- Reduction of quadratic form to canonical form — Rank — Positive, negative and semi definite — Index — Signature. Free vibration of a two-mass system. Curve tracing: Cartesian, Polar and Parametric forms.
Multiple integrals: Double and triple integrals — Change of variables —Change of order of integration. Finding Areas and Volumes. Evaluation of integrals. Gradient- Divergence- Curl — Laplacian and second order operators -Vector identities. Equation of continuity, potential surfaces. Line integral — Work done — Potential function — Area- Surface and volume integrals Vector integral theorems: Greens, Stokes and Gauss Divergence theorems without proof and related problems.
Work is done, Force. The Main Unit of the book are: Share this article with your classmates and friends so that they can also follow Latest Study Materials and Notes on Engineering Subjects. I want to download m3 material but not avalible plz send to my email.
Leave A Reply Cancel Reply. Save my name, email, and website in this browser for the next time I comment. June Solution: For this equation.. Therefore C. Using equation i.. Using this condition in expressions iv and v. The given equations can be written as D 7 x y Engineering Mathematics II 15MAT21 1 6t e c1 sin t c2 cos t 6e6t c1 cos t c2 sin t 5e6t c1 cos t c2 sin t 2 1 x c1 c2 e6t cos t c2 c1 e6t sin t Using this condition in expressions 3 and 4.
The given equations can be written as Dx 2 y cos 2t. Dec Solution: Dec e3 x Solution: Given Dx 2 y sin t.. II dx dy 1 Solve the simultaneous equations dt 2 y sin t 0. June Sol: The given equation can be written as p x tan 1 p. June Put D the given equations becomes dt Dx 7 x y 0 ; Dy 2 x 5 y 0. Jan Sol: The given equation is solvable for y only.
Jan Solution: The given equation is solvable for x only. Jan substitution X x2.
The given equation is. The given equation is solvable for x and it can be written as y 2x py. Thus the general solution is given by 2 y x 2 c e x y x 1 c 0 16 Obtain the general solution and singular solution of the equation y 2 px p2 y. The given equation can be written as xp 3 1 1 y 2 y px i s in theClairaut ' s form y px f p p p2 whose general solution is y cx f c 1 Thus general solution is y cx 2 c Differnetiating partially w.
Dividing throughout by p2. Jan z Sol: Jan Derive one dimensional heat equation as c. July t x2 Sol: Consider a heat conducting homogeneous rod of length L placed along x-axis. Department of Mathematics.
The rod is sufficiently thin so that the temperature is same at all points of any cross sectional area of the rod Also assume that the rod is insulated laterally and therefore heat flows only in the x direction. Therefore q1 the quantity of heat flowing into the cross section at a distance x in unit time is u q1 kA per second x x Negative sign appears because heat flows in the direction of decreasing temperature as x increases u decreases q2 the quantity of heat flowing out of the cross section at a distance x x i.
The amount of heat crossing any section of the rod per second depends on the area A of the cross u section Let u x.. Department of Mathematics.. July Sol: Equation 3 is the one dimensional heat equation which is second order homogenous and parabolic type.. Since the string does not offer resistance to bending.
The motion takes place entirely in the X Y plane. We shall find the displacement y a function of the distance x and the time t. Since the is no motion in the horizontal direction. Consider the motion of an element PQ of length s. Gravitational forces on the string are neglected. We shall obtain the equation of motion of string under the following assumptions. The string is perfectly flexible and offers no resistance to bending Points on the string move only in the vertical direction.
Consider a tightly stretched elastic string of length l stretched between two points O and A and displaced slightly from its equilibrium position OA. Let m be the mass per unit length of the string. Jan 2 Solution: Here we first find z by integration and apply the given conditions to determine the arbitrary functions occurring as constants of integration.
Equation 3 is the one dimensional wave equation which is second order homogenous and parabolic type Mass of an element PQ is m s..
The auxiliary equations are dx dy dz 2 2 2 x y z 2 xy 2 xz Taking the second and third terms we have. The auxiliary equations are dx dy dz 2 2 2 x yz y zx z xy Equivalently we can write in the form By data. June u u Solution: Let us suppose that z is a function of x only. Evaluate x2 y2 z 2 dxdydz Jan c b a c b a 2 2 z3 Solution: Evaluate xy dy dx changing the order of integration.
July 0 x2 12 x Sol: Let I xy dy dx 0 x 2 To identify the region of integration R let us find the point of intersection of the curves y x 2 and y 2 x x 1. July when y is an odd multiple of 2 z Sol: Sinx sin y. Evaluate x y z dy dx dz.