Leader: Course: Instructor: Date: Exam 1 Review 10am Section Supplemental Instruction Iowa State University Tess Math 267 Basnet 2/9/2016 Exam 1 Information: Molecular Biology 1414 10-10:50am. Sections Covered: 1.1-1.3, 2.1-2.5, 3.1-3. Integral Review Integral Solution Integral 1 ∫ 2 ππ’ π’ + π2 Solution 1 π‘ππ−1 π’ + πΆ π ∫ π π’ ππ’ ππ’ + πΆ 1 ∫ ππ’ π’ lnβ‘|π’| + πΆ ∫ ππ ππ’β‘πππ‘π’β‘ππ’ −cscu + πΆ ∫ πππ’β‘ππ’ π’β‘πππ’ − π’ + πΆ ∫ π πππ’β‘ππ’ ln(π πππ’ + π‘πππ’) + πΆ ∫ ππ’ ππ’ 1 π’ π +πΆ πππ ∫ π πππ’β‘ππ’ −πππ π’ + πΆ ∫ π ππ 2 π’β‘ππ’ π‘πππ’ + πΆ ∫ 1 √π2 − π’2 ππ’ ∫ π πππ’β‘ππ’ ∫ π’π ππ’ π’ +πΆ π −πππ π’ + πΆ π ππ−1 π’π+1 +πΆ π+1 Additional Methods: 2π₯ ο· U-Substitution → find u and du and sub into equation Ex) ∫ π₯ 2 +1 ππ₯ ο· Integration by Parts → ∫ π’ππ£ = π’π£ − β‘ ∫ π£ππ’ Ex) ∫ π₯π π₯ ππ₯ Method Review Method Separable 2.2 Linear (Standard) 2.3 Exact 2.4 Integration Factor 2.4 Bernoulli’s 2.5 General Form ππ¦ = π(π₯)β(π¦) ππ₯ ππ¦ + π(π₯)π¦ = π(π₯) ππ₯ π(π₯, π¦)ππ₯ + π(π₯, π¦)ππ¦ = 0 ππ¦ − ππ₯ ππ₯ − ππ¦ β‘ππβ‘ π π ππ¦ + π(π₯)π¦ = π(π₯)π¦ π ππ₯ Example DE ππ¦ π¦π π₯ = π −π¦ + π −2π₯−π¦ ππ₯ ππ¦ (π₯ + 1) = π π₯ (1 + π¦) ππ₯ 3 ππ¦ 3 (1 + + π₯) +π¦ = −1 π¦ ππ₯ π₯ π¦(π₯ + π¦ + 1)ππ₯ + (π₯ + 2π¦)ππ¦ = 0 ππ¦ − π¦ = π π₯π¦2 ππ₯ 1060 Hixson-Lied Student Success Center οΆ 515-294-6624 οΆ sistaff@iastate.edu οΆ http://www.si.iastate.edu 1. Solve the following IVP: π₯ ππ¦ + π¦ = π¦ 2 π₯ 2 πππ₯; β‘β‘β‘β‘β‘π¦(1) = 4 ππ₯ 2. Determine if the following DE is exact, if not determine the integrating factor and solve for general solution: π₯ 2 π¦β‘ππ₯ + π¦(π₯ 3 + π 3π¦ )ππ¦ = 0 3. Consider the following autonomous DE: a. Determine the critical Points ππ ππ₯ = −10 + π2 + 3π b. Create a phase porrate and classify points as stable, unstable or semi-stable 4. A small metal bar, whose initial tempeature was 20β, is dropped into a large container of boiling water. a. How long will is take the bar to reach 90β if it is known that its temperature increases 2β in 1 second? 5. Solve the following DE and give the general solution: ππ¦ 3π¦π₯ 2 (1 + 3π¦π ππ 2 π¦ + 4π¦ 4 ) + 3 = 0β‘ ππ₯ (π₯ + 2)