4 y 2 4 2 2y
[PDF File]22M:034 Engineer Math IV: Differential Equations
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2. Find the general solution of the nonhomogeneous equation y′′ − 2y′ − 3y = 3te2t. Solution. Wefirst findtherootsr1 = 3, r2 = −1ofthecharacteristic equation r2−2r−3 = 0. The general solution of the corresponding homogeneous equation is
[PDF File]Math V1202. Calculus IV, Section 004, Spring 2007 ...
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cylinder x2 +y2 = 4, oriented clockwise when viewed from above. Solution: Let S be the part of the plane 3x + 2y + z = 6 that lies inside the cylinder x 2 + y 2 = 1, oriented downward.
[PDF File]Solutions to Review Questions: Exam 3
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Therefore, Y(s) = G(s) 4s2 + 4s+ 17 = G(s) 1 8 2 (s+ 1 2)2 + 22 Therefore, y(t) = g(t) 1 8 e t=2 sin(2t) (b) y00+ y0+ 5 4 y= 1 u ˇ(t), with y(0) = 1 and y0(0) = 1. SOLUTION: Take the Laplace transform of both sides: (s2Y s+ 1) + (sY 1) 5 4 Y = 1 s e ˇs s so that
[PDF File]Maximum and Minimum Values
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• y =(1/4)x2 • x = 2y2 Next, we solve the system of equations. y =(1/4)(2y2)2 =) y = y4 If y =1,thenx = 2. If y =0,thenx =0. Thecriticalpoints (2,1) and (0,0). We now calculate the second derivatives to classify the critical point. • fxx(x,y)=6x • fyy(x,y)=48y • fxy(x,y)=12 138 of 155
[PDF File]Review for Exam 2. Section 14 - Michigan State University
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4 y2 in the closed triangular region with vertices given by (0,0), (1,0), and (0,2). Solution: I The horizontal side of the triangle, y = 0, x ∈ (0,1). Since g(x) = f (x,0) = 2 − 2x, ⇒ g0(x) = −2 6= 0 . there are no candidates in this part of the boundary. I The vertical side of the triangle is x = 0, y ∈ (0,2). Then, g(y) = f (0,y ...
[PDF File]Partial Differentiation
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EXAMPLE 14.1.2 We have seen that x2 +y2 +z2 = 4 represents a sphere of radius 2. We cannot write this in the form f(x,y), since for each x and y in the disk x 2 +y 2 < 4 there are two corresponding points on the sphere.
[PDF File]Lagrange Multipliers
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•Example 4 •Find the points on the sphere x2+y2+z2=4 that are closest to and farthest from the point (3,1,-1). •Solution: The distance from a point (x,y,z) to the point (3,1,-1) is d= (x−3)2+(y−1)2+(z+1)2 But the algebra is simple if we instead maximize and minimize the square of the distance: 2 d =f(x,y,z)=(x-3)2+(y-1)2+(z+1)2
[PDF File]Math 2263 Quiz 10 - University of Minnesota
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Math 2263 Quiz 10 26 April, 2012 Name: 1. Evaluate RR S zdS, where S is the part of the plane 2x+ 2y + z = 4 that lies in the rst octant. Answer: The x-, y-, and z-intercepts of the given plane are 2, 2, and 4.
[PDF File]Unit #5 - Implicit Di erentiation, Related Rates Implicit ...
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2x+ 2 2y 4 2x+ 2 = 0 x= 1 Substituting x= 1 back into the circle equation to nd the matching ycoordinates gives (1) 2+ y2(1) 4 = 1 y2 4y= 0 y(y 4) = 0 y= 0; 4 The points with horizontal tangents are (1,0) and (1,4). (b)The points with vertical tangents are those where the denominator of dy dx is zero (making the slope unde ned). From part (a ...
[PDF File]MATH 3321 Sample Questions for Exam 2 Second Order ...
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MATH 3321 Sample Questions for Exam 2 Second Order Nonhomogeneous Differential Equations: Section 3.4, 3.5 1. z1(x) = 2x3 + xln x, z2(x) = xln x − x3 are solutions of a second order, linear nonhomo- geneous equation L[y] = f(x). y1(x) = x−2 is a solution of the corresponding reduced equation L[y] = 0.
[PDF File]5.8 Lagrange Multipliers
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Multivariate Calculus; Fall 2013 S. Jamshidi 4. x4 +y4 +z4 =1 If x,y,z are nonzero, then we can consider Therefore, we have the following equations: 1. 1=2x2 2. 1=2y2 3. 1=2z2 4. x4 +y4 +z4 =1 Remember, we can only make this simplification if all the variables are nonzero!
[PDF File]Solution - University of Toledo
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cylinder z = (4−y2)1/2 below by the xy plane and the projection D of the solid onto the xy-plane is the triangle with edges x = 2y, x = 0 and the intersection of the cylinder with the plane z = 0 which gives y 2 = 4 or y = 2 (first octant).
[PDF File]Simultaneous Equations - YMLearn
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Step 2 Make the Solve the simultaneous equations 3x + 2y = 4 4x + 5y = 17 Step 1 – label each equation as equation 1 and equation 2 3x + 2y = 4 --‘> equation 1 4x + 5y = 17 –‘> equation 2 coefficient’s of x or y equal to each other 5 x equation 1 ---‘> 5 (3x + 2y = 4) 2 x equation 2 ---‘> 2(4x + 5y = 17)
[PDF File]2. Partial Differentiation - MIT OpenCourseWare
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2. PARTIAL DIFFERENTIATION 1 b) wx = −y2/x2, wy = 2y/x; therefore at (1,2,4), we get wx = −4, wy = 4, so that the tangent plane is w = 4−4(x −1)+4(y −2), or w = −4x +4y. x x y 2B-2 a) zx = = ; by symmetry (interchanging x and y), zy = ; then the x2 + y2 z z tangent plane is z = z0 0+ x0 (x−x y0 x0 x+ y0 y , since x2+y2 = z2. 0)+ (y−y0), or z =
[PDF File]Stokes’ Theorem
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(2)To come up with this parameterization, rewrite x2 + 4 y2 = 4 as x 2 2 + y2 = 1 and then use x 2 = cos t, = sin . It’s easy to check that it’s reasonable: if we plug in x = 2cost, y = sint, and z = 3, then the equations x2 + 4y2 = 4 and z = 3 are indeed satis ed. 2
[PDF File]Double integrals
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−2y + y3 2 # y=1 y=0 = 5 2 −2+ 1 2 = 1 5. Method 2 : do the integration with respect to y first and then x. In this approach we select a “typical x” and draw a vertical line across the region D at that value of x. Vertical line enters D at y = 0 and leaves at y = x. We then need to let x
[PDF File]Quadric Surfaces - CoAS
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After completing the square, we can rewrite the equation as: 4(x 21)2 + (y + 5) + 16(z + 1)2 = 37 This is a hyperboloid of 1 sheet which has been shifted. Speci cally, its central
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Name: SOLUTIONS Date: 10/06/2016 4x2 2(16 4x) + 1 = 8x2 15 for 2 x 2. The critical number here satis es 16x= 0 =)x= 0. With x= 0, y2 = 16 4 02 =)y= 4. So, the critical points on the boundary are (0; 4) and (0;4).
[PDF File]FINALEXAMPRACTICEPROBLEMS
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In fact, as a particular solution of y′′ +2y′ +y= e−t 9t we can take 1 9ln(t)te−9t because the term −1 9te −t is included in the solution space of the corresponding homogeneous equation. It follows that the general solution of the differential equation y′′ +2y′ +y= e−t 9t (t > 0) is y(t) = C 1e−t +C 2te−t + 1 9 ln(t)te−t where C
[PDF File]Separable Di fferential Equations
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4 y-2 -1 1 2t Figure 4: Solution of dy/dt = −2y, y(−1) = 2 In-class Exercise 3: Find the solution of the initial value problem dy dt =2y y(1) = 1 and sketch its graph. To be general, let us solve the initial value problem dy dt = ky (11) y(t0)=y0 (12) where k, t0,andy0 aregivenconstants: Thegeneralsolutionofthedi fferential equation (11 ...
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