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Emat 213 midterm 2 winter 2006

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The document contains a midterm exam for an ODE class with 6 problems worth 10 points each. Problem 1 asks to find the general solution of a 7th order linear ODE using the method of undetermined coefficients. Problem 2 asks to solve a 2nd order linear ODE using either variation of parameters or undetermined coefficients. Problem 3 asks to solve a nonlinear 2nd order ODE using a substitution. Problem 4 asks to find the equation of motion for a mass attached to a spring with an external force applied. Problem 5 asks to solve an eigenvalue problem for a CE equation. Problem 6 asks to use variation of parameters to solve a 2nd order nonhomogeneous ODE.

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Midterm Exam Emat 213 March 2006 The best four problems will be marked [10 points] Problem 1. Find the general solution (complementary + particular) of the linear ODE (for yp use the method of Undetermined Coefficients) y (7) − 6y (6) + 14y (5) − 20y (4) + 25y ′′′ − 22y ′′ + 12y ′ − 8y = x2 You may find this useful: m7 − 6m6 + 14m5 − 20m4 + 25m3 − 22m2 + 12m − 8 = (m − 2)3 (m + i)2 (m − i)2 Solution From the multiplicities of the roots yc = (c1 + c2 x + c3 x2 )e2x + (c4 + c5 x) cos(x) + (c6 + c7 x) sin(x) For yp we use “Undetermined Coefficients” yp = Ax2 + Bx + C ′ yp = 2Ax + B ′′ yp = 2A ′′′ yp = ... = 0 −44A + 24Ax + 12B − 8Ax2 − 8Bx − 8C = x2 1 A=− 8 3 −3 − 8B = 0 ⇒ B = − 8 44 36 1 − − 8C = 0 ⇒ C = 8 8 8 [10 points] Problem 2. Find the general solution (complementary + particular) of the linear ODE 1 y ′′ − y ′ + y = 3 + ex 4 [Hint: for yp you can use variation of parameters or undetermined coefficients, the latter is faster but it’s your choice]. Solution 1 m2 − m + = (m − 1/2)2 4 yc = (c1 + xc2 )e−x/2 yp = A + Bex 1 A (B − B + B)ex + = 3 + ex 4 4 B = 4 , A = 12 y = (c1 + xc2 )e−x/2 + 4ex + 12 1
[10 points] Problem 3. Solve the following nonlinear second order ODE by the substitution u = y ′ u du = y ′′ dy y ′′ + 2y(y ′ )3 = 0 Solution: problem 7 pag. 146. [10 points] Problem 4. When a mass of 2 Kilograms is attached to a spring whose constant is 32 N/mm, it comes to rest in the equilibrium position. Starting at t = 0, a force equal to Fext (t) = e−2t is applied to the system. Find the equation of motion in absence of damping. Solution 2x′′ + 32x = e−2t x(0) = 0, x′ (0) = 0 xc (t) = c1 cos(4t) + c2 sin(4t) xp = Ae−2t 8A + 32A = 1 1 A= 40 1 x(t) = c1 cos(4t) + c2 sin(4t) + e−2t 40 1 1 x(0) = c1 + = 0 ⇒ c1 = − 0 40 4 ′ 1 1 x (0) = 4c2 − = 0 ⇒ c2 = 20 80 So the solution of the motion is 1 1 1 x(t) = − cos(4t) + sin(4t) + e−2t 40 80 40 [10points] Problem 5. Solve the eigenvalue problem for the CE equation below with the prescribed boundary value conditions x2 y ′′ + xy ′ + λy = 0 y ′ (1) = 0 , y ′ (e2 ) = 0 Solution: problem 21 pag. 169 from the book. [10 pts] Problem 6. Using the method of variation of parameters find the general solution (complementary + particular) of the ODE √ 2y ′′ + 2y ′ + y = 4 x (x > 0) Note: the antiderivatives for u1 , u2 are transcendental (cannot be computed explicitly). Leave them indicated. Solution Problem 16 pag. 136. 2

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Emat 213 midterm 2 winter 2006

  • 1. Midterm Exam Emat 213 March 2006 The best four problems will be marked [10 points] Problem 1. Find the general solution (complementary + particular) of the linear ODE (for yp use the method of Undetermined Coefficients) y (7) − 6y (6) + 14y (5) − 20y (4) + 25y ′′′ − 22y ′′ + 12y ′ − 8y = x2 You may find this useful: m7 − 6m6 + 14m5 − 20m4 + 25m3 − 22m2 + 12m − 8 = (m − 2)3 (m + i)2 (m − i)2 Solution From the multiplicities of the roots yc = (c1 + c2 x + c3 x2 )e2x + (c4 + c5 x) cos(x) + (c6 + c7 x) sin(x) For yp we use “Undetermined Coefficients” yp = Ax2 + Bx + C ′ yp = 2Ax + B ′′ yp = 2A ′′′ yp = ... = 0 −44A + 24Ax + 12B − 8Ax2 − 8Bx − 8C = x2 1 A=− 8 3 −3 − 8B = 0 ⇒ B = − 8 44 36 1 − − 8C = 0 ⇒ C = 8 8 8 [10 points] Problem 2. Find the general solution (complementary + particular) of the linear ODE 1 y ′′ − y ′ + y = 3 + ex 4 [Hint: for yp you can use variation of parameters or undetermined coefficients, the latter is faster but it’s your choice]. Solution 1 m2 − m + = (m − 1/2)2 4 yc = (c1 + xc2 )e−x/2 yp = A + Bex 1 A (B − B + B)ex + = 3 + ex 4 4 B = 4 , A = 12 y = (c1 + xc2 )e−x/2 + 4ex + 12 1
  • 2. [10 points] Problem 3. Solve the following nonlinear second order ODE by the substitution u = y ′ u du = y ′′ dy y ′′ + 2y(y ′ )3 = 0 Solution: problem 7 pag. 146. [10 points] Problem 4. When a mass of 2 Kilograms is attached to a spring whose constant is 32 N/mm, it comes to rest in the equilibrium position. Starting at t = 0, a force equal to Fext (t) = e−2t is applied to the system. Find the equation of motion in absence of damping. Solution 2x′′ + 32x = e−2t x(0) = 0, x′ (0) = 0 xc (t) = c1 cos(4t) + c2 sin(4t) xp = Ae−2t 8A + 32A = 1 1 A= 40 1 x(t) = c1 cos(4t) + c2 sin(4t) + e−2t 40 1 1 x(0) = c1 + = 0 ⇒ c1 = − 0 40 4 ′ 1 1 x (0) = 4c2 − = 0 ⇒ c2 = 20 80 So the solution of the motion is 1 1 1 x(t) = − cos(4t) + sin(4t) + e−2t 40 80 40 [10points] Problem 5. Solve the eigenvalue problem for the CE equation below with the prescribed boundary value conditions x2 y ′′ + xy ′ + λy = 0 y ′ (1) = 0 , y ′ (e2 ) = 0 Solution: problem 21 pag. 169 from the book. [10 pts] Problem 6. Using the method of variation of parameters find the general solution (complementary + particular) of the ODE √ 2y ′′ + 2y ′ + y = 4 x (x > 0) Note: the antiderivatives for u1 , u2 are transcendental (cannot be computed explicitly). Leave them indicated. Solution Problem 16 pag. 136. 2