Assignemnt3Sol

Assignemnt3Sol - Muhammad Ulhaque due at 11:55 PM...

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Unformatted text preview: Muhammad Ulhaque due 2/10/08 at 11:55 PM. Assignment 3 MATH263, Winter 2008 with initial conditions y1 0 1 y1 0 0 y1 Find the function y2 of t which is the solution of with initial conditions y2 Find the Wronskian 2.(1 pt) Solve the initial value problem 3.(1 pt) Solve the boundary value problem d2y yx = . Correct Answers: (0)*exp(3*x) + (0.00571894068498315)*x*exp(3*x) 8.(1 pt) Suppose that the second order linear equation 4.(1 pt) Find a fundamental set of solutions y1 x y2 x of the equation d 2y dy x2 2 9x 26y 0 dx dx in the interval x 0. y1 x = . y2 x = . Correct Answers: a*(x**(5))*cos(1*ln(x)) + b*(x**(5))*sin(1*ln(x)) 9.(1 pt) Let a be a real constant. Consider the equation a*(x**(5))*cos(1*ln(x)) + b*(x**(5))*sin(1*ln(x)) d2y dy 2 ay 0 dx2 dx 5.(1 pt) Solve the initial value problem with boundary conditions y 0 0 and y 4 0. 2 dy 2d y For certain discrete values of a, this equation can have non-zero x 9x 25y 0 y1 5y 1 28 dx2 dx solutions. Find the three smallest values of a for which this is yx = . the case. Correct Answers: Enter your answers in increasing order. , a2 , a3 a1 -5*x**(5) + -3*(x**(5))*ln(x) Correct Answers: 6.(1 pt) Solve the boundary value problem 1.61685027506808 2 dy 2d y 3.46740110027234 x 5x 5y 0 y1 0y2 1 dx2 dx 6.55165247561276 yx = . 10.(1 pt) Match the third order linear equations with their Correct Answers: fundamental solution sets. (0.0333333333333333)*x**(5) + (-0.0333333333333333)*x**(1)2y 1. y y 2y 0 1 d 2y dy 4x 3x3 2y 0 dx2 dx has a fundamental set of solutions y1 y2 such that y1 1 1, y1 1 2, y2 1 5 and y2 1 3. Then find explicitly the Wronskian w x y 1 x y2 x y2 x y1 x in x 0. We have w x Correct Answers: -13*x**(--4)*exp(0.5*-3*(1-x*x)) x2 ! dx2 6 9y 0 y0 0y 2 1 dy dx Correct Answers: (1/2 - -5/16) *exp((5/9 - 8/9)*t) + (1/2 + -5/16) *ex (0 - 9/16) *exp((5/9 - 8/9)*t) + (0 + 9/16) *exp((5/9 exp(--90/81 *t) . yx = Correct Answers: 1*exp(3*x) + -4*exp(1*x) dy 4 dx 3y 0 y0 3y 0 1 W t Remark: You can find W by direct computation and use Abel's theorem as a check. You should find that W is not zero and so y1 and y2 form a fundamental set of solutions of 81y 90y 39y 0 d2y dx2 W t W y 1 y2 y2 0 0 y2 0 81y 90y 39y 0 1 1.(1 pt) Find a fundamental set of solutions y1 x y2 x of the equation d2y dy 6 10y 0 dx2 dx y1 x = . y2 x = . Correct Answers: a*exp(3*x)*cos(1*x) + b*exp(3*x)*sin(1*x) a*exp(3*x)*cos(1*x) + b*exp(3*x)*sin(1*x) 7.(1 pt) Find the function y1 of t which is the solution of 81y 90y 39y 0 Prepared by the WeBWorK group, Dept. of Mathematics, University of Rochester, c UR 2 A. B. C. D. 1 cos t sin t 1 e5t e4t e t te t t 2 e t et tet e t # " " " " 2. 3. 4. 5. 6. y y y ty y 9y 20y 0 y 0 y y y 0 y 0 3y 3y y 0 E. 1 t t 3 F. e2t cos t sin t Correct Answers: F B A D E C ...
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This note was uploaded on 06/09/2008 for the course MATH MATH 263 taught by Professor Humphire during the Spring '08 term at McGill.

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