HW2solutions - Differential Equations Assignment 2:...

Info iconThis preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Differential Equations Assignment 2: (Solutions) Quantitative Methods for First Order Equations and Modeling: 1. Use the method of integrating factors to find the general solution to the following first order differential equations: (a) y + 3 y = 0 The integrating factor is given by = e 3 t . Multiplying by this factor we find: e 3 t y + 3 e 3 t y = 0 . Applying the product rule in reverse we know: d dt e 3 t y = 0 . Integrating with respect to t and solving for y we obtain the general solution. Z d dt e 3 t y dt = Z dt, e 3 t y = C, y = Ce- 3 t . (b) ty + 2 y = 1 t First we write the equation in the proper form for applying the integrating factor formula y + 2 t y = 1 t 2 Now we find the integrating factor. = e R 2 t dt = e 2 ln( t ) = e ln( t 2 ) = t 2 . Multiplying by this integrating factor: t 2 y + 2 ty = 1 . 1 Applying the product rule in reverse we know: d dt t 2 y = 1 . Integrating with respect to t and solving for y we obtain the general solution. Z d dt t 2 y dt = Z 1 dt, t 2 y = t + C, y = 1 t + C t 2 . (c) y- y = e- t The integrating factor for this equation is = e- t . Multiplying by this factor we obtain: e- t y- e- t y = e- 2 t . Using the product rule in reverse we know: d dt e- t y = e- 2 t . Integrating with respect to t and solving for y we obtain the general solution. Z d dt e- t y dt = Z e- 2 t dt, e- t y =- 1 2 e- 2 t + C, y =- 1 2 e- t + Ce t . (d) y + cot( t ) y = 0 For this problem, we sidestep the calculation of the integrating factor and take advantage of the structure of the equation. Using the definition of cot( t ) we know: y + cos( t ) sin( t ) y = 0 2 Multiplying by sin( t ) , we find that we already have the correct form to apply the method of integrating factors: sin( t ) y + cos( t ) y = 0 d dt [sin( t ) y ] = 0 Integrating with respect to t and solving for y we obtain the general solution. Z d dt [sin( t ) y ] dt = 0 sin( t ) y = C y = C sin( t ) = C csc( t ) 2. Solve the following initial value problems using any appropriate technique: (a) y + 4 y = e- 2 t , y (0) = 1, This is a linear equation, so the method of integrating factors will work. The integrating factor is = e 4 t . Multiplying by this factor we get: e 4 t y + 4 e 4 t y = e 2 t Applying the product rule in reverse we find: d dt e 4 t y = e 2 t . Integrating and solving we get: Z d dt e 4 t y dt = Z e 2 t dt. e 4 t y = 1 2 e 2 t + C y = 1 2 e- 2 t + Ce- 4 t Substituting the initial condition we find: 1 = 1 2 + C C = 1 2 3 Thus, the particular solution to this initial value problem becomes: y = 1 2 e- 2 t + 1 2 e- 4 t Since this solution contains only exponentials of the form e- at , each term will go to zero and y and t ....
View Full Document

Page1 / 11

HW2solutions - Differential Equations Assignment 2:...

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online