math-33b-midterm-1

Math 33B Midterm 1 Differential Equations (University of California Los Angeles) Scan to open on Studocu Studocu is not sponsored or endorsed by any college or university Math 33B Midterm 1 Differential Equations (University of California Los Angeles) Scan to open on Studocu Studocu is not sponsored or endorsed by any college or university Downloaded by mingus bai (mingusbai@gmail.com) lOMoARcPSD|22977868

22F-MATH-33B-LEC-2 Midterm 1 KATHERINE CALLAHAN TOTAL POINTS 31.25 / 45 QUESTION 1 Problem 1 10 pts 1.1 1a 5 / 7 ✓ + 1 pts Noting this is a separable equation or separating variables to get $$\int \frac{1}{y^3}dy = \int \frac{t}{\sqrt{1+t^2}}dt $$ + 2 pts Integrating properly, to get $$\frac{-1}{2y^2} = \sqrt{1+t^2} +C$$, ✓ + 1 pts Plugging in Initial values, ie getting $$\frac{- 1}{2 (1^2)} = \sqrt{1+0^2} +C$$ or the appropriate equation based on earlier work ✓ + 1 pts Solving for $$C$$. + 2 pts Proving an explicit solution for $$y$$ given the work above. I.e. if all else is correct writing $$y(t) = \frac{1}{\sqrt{3-2\sqrt{1+t^2}}}$$ ✓ + 1 pts Giving an explicit solution for $$y$$ but the work to get it from the implicit definition had some errors + 0.5 pts Giving only an implicit definition of $$y$$ that solves the initial value problem, ie $$\frac{- 1}{2y^2} = \sqrt{1+t^2}-\frac{3}{2}$$ (this also includes writing plus or minus) + 0 pts blank, no progress ✓ + 1 pts Small errors in integrating 1.2 1b 1 / 3 ✓ - 1 pts Upper/lower bound is correct but the other is not ✓ - 1 pts Error in computing the bounds QUESTION 2 Problem 2 10 pts 2.1 2a 2 / 2 ✓ - 0 pts Correct 2.2 2b 5.5 / 8 ✓ - 2.5 pts Need to give an equation that defines solution implicitly. You have found $$F(x,y)$$ only. Answer: $$x^3-x^2y+2x+2y^3+3y=C$$ QUESTION 3 3 Problem 3 8 / 10 ✓ - 2 pts $$x(t)$$ is found incorrectly (after the integrating factor is found). QUESTION 4 Problem 4 8 pts 4.1 4a 3.5 / 5 ✓ - 1.5 pts open circle or close circle for the equilibrium points 4.2 4b 3 / 3 ✓ - 0 pts Correct QUESTION 5 Problem 5 7 pts 5.1 5a 1.25 / 5 ✓ - 0 pts Need to mention that there is no rectangle R that contains the initial point (1,0) where both $$f$$ and $$\frac{\partial f}{\partial y}$$ are continuous. ✓ - 3.75 pts Incorrect application of uniqueness theorem / incorrect hypotheses. Note: uniqueness is not guaranteed. 1 $$f(t,y)$$ 2 $$\frac{\partial f}{\partial y}$$ is incorrect 3 this is not one of the conditions of uniqueness theorem Downloaded by mingus bai (mingusbai@gmail.com) lOMoARcPSD|22977868

5.2 5b 2 / 2 ✓ - 0 pts Correct (C) Page 2 Downloaded by mingus bai (mingusbai@gmail.com) lOMoARcPSD|22977868

Downloaded by mingus bai (mingusbai@gmail.com) lOMoARcPSD|22977868