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### hw2key

Course: MATH 124, Spring 2009
School: Washington
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351 AMATH Homework 2 Keys July 8, 2009 Section2.1 14,16,19,31,40 Section2.4 3,12,15,29 Section2.6 8,21,23,24 Section 2.1 In each of Problems 13 through 20 nd the solution of the given initial value problem. 14. y + 2y = te2t , y (1) = 0 Multiply (t) = e2t , then e2t y + 2e2t y ey e2t y y 2t =t =t = t2 +C 2 t2 +C = e2t 2 Apply initial condition y (1) = So the solution is 1 2 1 + C e2 = 0 = C = 2 . y =...

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351 AMATH Homework 2 Keys July 8, 2009 Section2.1 14,16,19,31,40 Section2.4 3,12,15,29 Section2.6 8,21,23,24 Section 2.1 In each of Problems 13 through 20 nd the solution of the given initial value problem. 14. y + 2y = te2t , y (1) = 0 Multiply (t) = e2t , then e2t y + 2e2t y ey e2t y y 2t =t =t = t2 +C 2 t2 +C = e2t 2 Apply initial condition y (1) = So the solution is 1 2 1 + C e2 = 0 = C = 2 . y = e2t t2 1 2 2 16. y + (2/t)y = (cos t)/t2, y ( ) = 0, t > 0 2 Integrating factor (t) = e t dt = t2 . Multiply it to the DE, we get y t2 + 2ty ty t y y 2 2 = = = = cos t cos t sin t + C t2 (sin t + C ) 1 Apply initial condition, we have y ( ) = 2 (sin + C ) = C 2 = 0 = C = 0 Plug C back into the general solution, y (t) = t2 sin t 19. t3 y + 4t2 y = et , y (1) = 0, t < 0 First write it in the standard form 4 y + y = t3 et t 4 (t) = e t dt = t4 then multiply it, t4 y + 4t3 y t4 y t4 y y = = = = tet te t tet dt = tet et + C e4 tet et + C Apply initial condition y (1) = e4 (e e + C ) = e4 C = 0 = C = 0 Plug it into the general solution, we get y (t) = e4t (t + 1) 31. Consider the initial value problem 3 y y = 3t + 2et , 2 y (0) = y0 . Find the value of y0 that separates solutions that grow positively as t from those that grow negatively. How does the solution that corresponds to this critical value of y0 behave as t ? Integrating factor 3 3 (t) = e 2 dt = e 2 t 2 So e 2 t y e 2 t y 3 3 3t + 2et e 2 t = 3te 2 t + 2e 2 3 t 2t = 3te dt + 2 e 2 dt 3 3 t = 2 te 2 t e 2 t dt 4e 2 = 3 3 t 3 t 43 = 2te 2 t e 2 t 4e 2 + C 3 3 4 y (t) = 2t 4et + Ce 2 t 3 4 Apply initial condition y (0) = 3 4 C = y0 = C = y0 + 16 . So the specic 3 solution is y (t) = 2t 4 16 3 4et + (y0 + )e 2 t 3 3 where the last term is the dominant term. Thus the critical value of y0 is the value that makes it 0, i.e. y0 + 16 16 = 0 = y0 = 3 3 when y0 > 16 , y (t) as t + 3 when y0 < 16 , y (t) + as t + 3 when y0 = 16 , y (t) = 2t 3 4 3 4et as t + In Problem 40 use the method of Problem 38 to solve the given dierential equation. 40. y + (1/t)y = 3 cos 2t, t>0 Based on 38, we want to nd out A(t) that satises A (t) A(t) 3 cos 2t e t dt = 3t cos 2t 3 3 = 3t cos 2tdt = t sin 2t + cos 2t + C 2 4 = 1 So the general solution is y (t) = A(t)e = 1 t dt = A(t) t 3 3 C sin 2t + cos 2t + 2 4t t 3 Section 2.4 In each of Problems 1 through 6 determine (without solving the problem) and interval in which the solution of the given initial value problem is certain to exist. 3. y + (tan t)y = sin t, y ( ) = 0 This is a rst order linear DE. p(t) = t tan t is continuous on k , k + 2 2 where k is an integer. q (t) = sin t is continuous on (, +). According to Thm 2.4.1, the solution to this IVP exists on , 32 . 2 In each of Problems 7 through 12 state where in the ty -plane the hypotheses of Teorem 2.4.2 are satised. 12. dy = (cot t)y dt 1+y Function f (t, y ) is continuous in (k, k + ) (, 1) and (k, k + ) (1, +). f cot t y = (1+y )2 , which is continuous in the same area as that of f (t, y ). So Thm 2.4.2 is satised when t = k, k Z and y = 1 In each of Problems 13 through 16 solve the given initial value problem determine and how the interval in which the solution exists depends on the initial value y0 . 15. y + y 3 = 0, y (0) = y0 y = y 3 This is a separable equation, we need to divide y 3 , then let's check the case of y (x) 0. Obviously, it is a solution. If y (t) = 0, 2 dy y3 y 2 y 2 = = = 2dx 2x + C 1 2x + C For the initial condition, If y0 > 0, 2 y0 = 1 1 C= 2 y= C y0 y0 2 2xy0 + 1 If y0 < 0, 2 y0 = 1 1 C= 2 y= C y0 1 2x + 1 2 y0 = 2 y0 2+1 = 2xy0 y0 2 2xy0 + 1 4 If y0 = 0, then y falls into the rst case we talked above: y (x) 0. Now let's analyze the interval on which the solution exists, 1 1 2 If y0 = 0, 2xy0 + 1 > 0 x > 2y2 2y2 , + ; 0 0 If y0 = 0, no ristriction on x (, +). Sometimes it is possible to solve a nonlinear equation by making a change of the dependent variable that converts it into a linear equation. The most important such equation has the form Bernoulli Equations. y + p(t)y = q (t)y n . and is called a Bernoulli equation after Jakob Bernoulli. Problem 27 through 31 deal with equations of this type. 29. y = ry ky 2 , r > 0 and k > 0. 1 Substitute v = y 1 , then y = v , dy = v12 dv . Now the DE becomes dt dt 1 dv 1 r v 2 dt v dv + rv dt = k =k 1 v2 Integrating factor (t) = ert , multiply it to the above equation, ert v v (t) y (t) = = = kert k + Cert r 1 r = v (t) k + Crert Section 2.6 Determin whether each of the equations in Problems 1 through 12 is exact. If it is exact, nd the solution. 8. (ex sin y + 3y )dx (3x ex sin y )dy = 0 Check P y Q Q x = ex cos y + 3 (3 + ex sin y ) ex cos y + 6 ex sin y = 3x + ex sin y 3x + ex sin y Q x Check P P y = ex cos y 6 + ex sin y ex sin y + 3y So it's not an exact equation. 5 Show that the equations in Problems 19 through 22 are not exact but become exact when multiplied by the given integrating factor. Then solve the equations. 21. ydx + (2x yey )dy = 0, (x, y ) = y Check Q P =1= =2 y x which is not exact. Multiply (x, y ) = y , the DE becomes y 2 dx + 2xy y 2 ey dy = 0 Check which means exact now.So P Q = 2y = = 2y y x x y = y2 = 2xy y 2 ey From the rst equation, we have (x, y ) = xy 2 + (y ) , then = 2xy + (y ) = 2xy y 2 ey y = (y ) = y 2 ey + 2yey 2ey So the nal solution is (x, y ) = xy 2 y 2 ey + 2yey 2ey = C 23. Show that if (Nx My )/M = Q, where Q is a function of y only, then the dierential equation M + Ny = 0 has an integrating factor of the form (y ) = exp Q(y )dy. If the integrating factor is (y ), we multiply it to the equation and get, M dx + N dy = 0 6 which now should be an exact equation, i.e. (M ) y d M M + dy y 1 d dy = = = (N ) x N x Nx M y =Q M Q(y )dy exp Q(y )dy ln || = (y ) = 24. Show that if (Nx My )/(xM yN ) = R, where R depends on the quantity xy only, then the dierential equation M + Ny = 0 has an integrating factor of the form (xy ). Find a general formula for this integrating factor. If the integrating factor is denoted by (xy ), multiply it to the DE we should get an exact equation, M dx + N dy = 0 then (M ) y M M + y x = = (N ) x N N + x y N M x y N y (xy ) M x (xy ) (yN xM ) M x Ny =R yN xM xy exp R (s) ds (Mx Ny ) = = = = = 7
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