MATH 285 E1 F1 GRADED HOMEWORK SET 1 DUE WEDNESDAY SEPTEMBER 10 IN LECTURE IT WOULD BE SO SWEET if you followed these instructions Please put each problem on a separate sheet of paper with your name and section E1 or F1 If a problem runs multiple pages please staple all the pages for a single problem together Think of each problem as a separate assignment This may be annoying but it will greatly streamline the grading process resulting in faster feedback for you Thank you Section and problem numbers refer to Differential Equations Boundary Value Problems Fourth Edition by Edwards and Penney 1 Let f x be the function defined piece wise as x if x 5 f x 5 if x 5 Find the solution of the initial value problem dy f x y 0 100 dx Hint Your solution will also be defined piece wise Solution 100 21 x2 if x 5 y x 25 100 2 5x if x 5 To obtain this first consider the range x 5 The equation becomes dy 1 2 dx x with general solution y x 2 x C In order to satisfy the initial condition y 0 100 the constant must be C 100 This gives the first part of the piece wise definition dy 5 Second consider the range x 5 The equation becomes dx with general solution y x 5x D We have to choose the value of D so that the two pieces match at x 5 that is so that y x is a continuous function At x 5 the formula 100 12 x2 has the value 25 25 100 25 2 So we need 5 5 D 100 2 hence D 100 2 This gives the second part of the piece wise definition 2 Consider the differential equation dy x dx y Sketch the slope field for this equation What are the solution curves Hint You should recognize them as semi familiar geometric shapes 1 2 MATH 285 E1 F1 GRADED HOMEWORK SET 1 DUE WEDNESDAY SEPTEMBER 10 IN LECTURE Solution The slope field is perpendicular to the lines through the origin The solution curves are the upper and lower half circles centered at the origin Note that an entire circle is not a solution curve because it does not define y as a function of x 3 Section 1 4 problem 22 Solution y x 3ex Starting from dy dx 4 x 4x3 y y separate variables to obtain Z Z dy 4x3 1 dx y ln y x4 x C y eC ex 4 4 4 x Dex where the constant D absorbs the plus minus sign and eC We now use the initial condition y 1 3 4 1 3 y 1 De1 De0 D So D 3 4 Section 1 5 problem 10 Find the general solution valid for x 0 Solution y x 3x3 Cx3 2 We first put the equation 2xy 0 3y 9x3 into standard form y0 9 3 y x2 2x 2 An integrating factor is R e 3 2x dx 3 e 2 ln x eln x 3 2 x 3 2 Since we are restricting the domain to x 0 we may simply use x 3 2 as the integrating factor Multiply through 3 9 x 3 2 y 0 x 5 2 y x1 2 2 2 d 3 2 9 x y x1 2 dx 2 Integrate x 3 2 y 3x3 2 C Solve for y y 3x3 Cx3 2 MATH 285 E1 F1 GRADED HOMEWORK SET 1DUE WEDNESDAY SEPTEMBER 10 IN LECTURE 3 5 Section 1 6 problem 14 Solution p y x 2Cx C 2 p Starting from yy 0 x x2 y 2 use the substitution u x2 y 2 Then u0 2x 2yy 0 which we recognize as 2 times the left hand side The equation becomes 1 2 u0 u Separate variables Z 1 2 u 1 2 du Z dx u1 2 x C u x C 2 Finally solve for y x2 y 2 u x C 2 q y x C 2 x2 p y 2Cx C 2
View Full Document
Unlocking...