MATH 202 Exam 1, Spring 2002

MATHEMATICS 202 SPRING SEMESTER 2006-2007 QUIZ I Time: 80 MINUTES. Date: March 26, 2007. Name:—————————————————————————- ID Number:———————————————————————– Section Number:—————————————————————— Course Instructors: Professors A. Lyzzaik and H. Yamani Question Grade 1 /14 2 /14 3 /14 4 /14 5 /14 6 /14 7 /18 TOTAL /100 Answer The Following Seven Questions On The Page Al- located For Each Question (You May Use The Back Of The Pages If Needed). 1 2 1. Show that the differential equation (−xy sin x + 2y cos x) dx + 2x cos x dy = 0 is not exact. Multiply the equation by an appropriate integrating factor µ(x, y) = xmyn that makes the differential equation exact, then solve. (14 points) Solution. Since (−xy sin x+2y cos x)y = −x sin x+2 cos x ̸= 2 cos x−2x sin x = (2x cos x)x, the differential equation is not exact. Multiply both sides of the differential equation by µ(x, y) = xmyn to obtain (−xm+1yn+1 sin x + 2xmyn+1 cos x) dx + 2xm+1yn cos x dy = 0. For this equation to be exact we must have (−xm+1yn+1 sin x + 2xmyn+1 cos x)y = (2xm+1yn cos x)x or −(n+1)xm+1yn sin x+2(n+1)xmyn cos x = 2(m+1)xmyn cos x−2xm+1yn sin x. This implies that m = n = 1. Hence the differential equation (−x2y2 sin x + 2xy2 cos x) dx + 2x2y cos x dy = 0 is exact. Let fx = −x2y2 sin x + 2xy2 cos x and fy = 2x2y cos x. Then f(x, y) = � 2x2y cos x dy = x2y2 cos x + g(x). Hence, 2xy2 cos x − x2y2 sin x + g′(x) = fx = −x2y2 sin x + 2xy2 cos x which yields g′(x) = 0. Hence, f(x, y) = x2y2 cos x and the general solution is x2y2 cos x = c for any constant c. 3 2. Solve the initial-value problem (14 points) x2 dy dx − 2xy = 3y4, y(1) = 1/2. Solution. Write the differential equation as dy dx − 2 xy = 3 x2y4. This it is a Bernoulli’s differential equation. Then write the differential equation as y−4 dy dx − 2 xy−3 = 3 x2. Letting u = y−3 yields du dx = −3y−4 dy dx or y−4 dy dx = −1 3 du dx. Then substitution in the latter differential equation yields −1 3 du dx − 2 xu = 3 x2, or du dx + 6 xu = − 9 x2. An integrating factor for this equation is µ(x) = e � (6/x)dx = x6. Multiplying the latter differential equation by µ yields x6du dx + 6x5u = −9x4 or d dx(x6u) = −9x4. Now integration of both sides gives x6u = −9 5x5 + c or y−3 = −9 5x−1 + cx−6. But y(1) = 1/2 gives 8 = −9/5 + c, or c = 9/5. Therefore, the solution for the IVP is y−3 = −9 5x−1 + 49 5 x−6. 4 3. Solve the initial-value problem (14 points) �y x − x2 y2 � dx − dy = 0, y(1) = 2. Solution. Write the differential equation as (y3 − x3)dx − xy2 dy = 0. Thus this is a homogeneous differential equation. If y = ux, then dy = u dx + x du. Then substitution in the latter differential equation yields x3(u3 − 1)dx − x3u2(u dx + x du) = 0, or (u3 − 1)dx − u2(u dx + x du) = 0, or dx + xu2 du = 0. Write this differential equation as u2 du + dx x = 0, then integrate both sides to obtain 1 3u3 + ln |x| = c, or y3 3x3 + ln |x| = c. But y(1) = 2 yields c = 8/3. Therefore, the solution for the IVP is y3 + 3x3 ln |x| = 8x3. 5 4. Solve the initial-value problem (14 points) dy dx = ex+y + e−x−y − 1, y(0) = 0. Determine the largest interval of definition of the solution. Solution. If u = x + y, then du dx = 1 + dy dx, or dy dx = du dx − 1. Then substitution in the differential equation yields du dx − 1 = eu + e−u − 1, or du dx = eu + e−u. Write this differential equation as eu e2u + 1du = dx, then integrate both sides to obtain arctan eu = x + c, or ex+y = tan(x + c). But y(0) = 0 yields c = π/4. Therefore, the solution for the IVP is ex+y = tan(x + π/4), or y = ln(tan(x + π/4)) − x. 6 5. Sketch the largest region in the xy−plane containing the points (x0, y0) for which the initial-value problem � 1 − y2 dy dx = √ 1 + x2, y(x0) = y0, possesses a unique solution; justify your answer. Justify your answer. (14 points) Solution. Write this differential equation as dy dx = √ 1 + x2 √1 − y2 , and let f(x, y) = √ 1 + x2 √1 − y2 . Since f and fy = y √ 1 + x2 (1 − y2)3/2 are continuous functions in the region R = {(x, y) : |y| < 1}, by the existence and uniqueness theorem, every initial-value problem � 1 − y2 dy dx = √ 1 + x2, y(x0) = y0, where (x0, y0) ∈ R possesses a unique solution. Moreover, since f is not defined whenever y = ±1, R is largest desired region. 7 6. Use Stoke’s theorem to evaluate the circulation of the vector field F(x, y, z) = −yi + x2j + z3k along the curve C which is the intersection of the circular cylinder x2 + y2 = 4 and the plane x+z = 3, and which is traversed counterclockwise when viewed from high up on the positive z−axis. (14 points) Solution. By Stoke’s theorem to evaluate the circulation of the given vector field is given by � C F · T ds = � � R ∇ × F · n dσ, where R is the disc x2 + y2 = 4, n and dσ are the unit surface normal and the surface area differential for the given surface. Thus, n = ∇(x + z − 3) |∇(x + z − 3)| = i + k √ 2 and dσ = |∇(x + z − 3)| |∇(x + z − 3) · k|dA = √ 2 dA. But ∇ × F = ������� i j k ∂/∂x ∂/∂y ∂/∂z −y x2 z3 ������� = (2x + 1)k Therefore, the desired circulation is � � S ∇ × F · n dσ = � � R(2x + 1)k · i + k √ 2 √ 2 dA = � � R(2x + 1)dA = � 2π 0 � 2 0 (2r cos θ + 1)r dr dθ = � 2π 0 [2r3 cos θ/3 + r2/2]2 0dθ = � 2π 0 [16 cos θ/3 + 2]dθ = [16 sin θ/3 + 2θ]2π 0 = 4π. 8 7. Evaluate in two different ways the flux of the vector field F(x, y, z) = xi + yj + zk across the surface S which is the portion of the paraboloid z = 4−x2−y2 above the xy−plane (n is upward-oriented so that its z−component is positive). (16 points) Solution. By taking z = 0 in the equation of the paraboloid, we conclude that the paraboloid meets the xy−plane in the circle x2+y2 = 4. The desired flux is given by � � S F · n dσ, where dσ is the surface area differential for the paraboloid and n = ∇(x2 + y2 + z − 4) |∇(x2 + y2 + z − 4)| = 2xi + 2yj + k √4x2 + 4y2 + 1. Thus the flux is � � S F · n dσ = � � S x2 + y2 + 4 √4x2 + 4y2 + 1 dσ. Now we find dσ in two ways: (a) dσ = |∇(x2 + y2 + z − 4)| |∇(x2 + y2 + z − 4) · k| dA = (4x2 + 4y2 + 1) dA. In this case, the flux is � � S F · n dσ = � � R(x2 + y2 + 4) dA = � 2π 0 � 2 0 (r2 + 4)r dr dθ = 16π. (b) dσ = |rr × rθ|drdθ, where r(r, θ) = (r cos θ)i + (r sin θ)j + (4 − r2)k. Then rr × rθ = ������� i j k cos θ sin θ −2r −r sin θ r cos θ 0 ������� , and rr × rθ = (2r2 cos θ)i + (2r2 sin θ)j + rk. Thus, dσ = √ 4r4 + r2 dr dθ = √ 4r2 + 1 rdr dθ. The rest of the work goes as in (a).