Mathematics 22: Lecture 7

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1 Mathematics 22: Lecture 7 Separation of Variables Dan Sloughter Furman University January 15, 2008 Dan Sloughter (Furman University) Mathematics 22: Lecture 7 January 15, / 8

2 Separable equations We call a differential equation of the form separable. du dt = g(t)h(u) Dan Sloughter (Furman University) Mathematics 22: Lecture 7 January 15, / 8

3 Separable equations We call a differential equation of the form du dt = g(t)h(u) separable. The general solution is 1 h(u) du = g(t)dt. Dan Sloughter (Furman University) Mathematics 22: Lecture 7 January 15, / 8

4 Consider the equation du dt = t u. Dan Sloughter (Furman University) Mathematics 22: Lecture 7 January 15, / 8

5 Consider the equation du dt = t u. Then udu = tdt. Dan Sloughter (Furman University) Mathematics 22: Lecture 7 January 15, / 8

6 Consider the equation du dt = t u. Then udu = tdt. Hence 1 2 u2 = 1 2 t2 + k for some constant k. Dan Sloughter (Furman University) Mathematics 22: Lecture 7 January 15, / 8

7 Consider the equation du dt = t u. Then udu = tdt. Hence 1 2 u2 = 1 2 t2 + k for some constant k. Or u 2 + t 2 = c for some constant c. Dan Sloughter (Furman University) Mathematics 22: Lecture 7 January 15, / 8

8 Consider the equation du dt = t u. Then udu = tdt. Hence 1 2 u2 = 1 2 t2 + k for some constant k. Or u 2 + t 2 = c for some constant c. Note: To solve for u as a function of t, we need to know which part of the circle u 2 + t 2 = c to use. Dan Sloughter (Furman University) Mathematics 22: Lecture 7 January 15, / 8

9 Consider the equation du dt = t u. Then udu = tdt. Hence 1 2 u2 = 1 2 t2 + k for some constant k. Or u 2 + t 2 = c for some constant c. Note: To solve for u as a function of t, we need to know which part of the circle u 2 + t 2 = c to use. For example, if u(1) = 3, then c = = 10. Dan Sloughter (Furman University) Mathematics 22: Lecture 7 January 15, / 8

10 Consider the equation du dt = t u. Then udu = tdt. Hence 1 2 u2 = 1 2 t2 + k for some constant k. Or u 2 + t 2 = c for some constant c. Note: To solve for u as a function of t, we need to know which part of the circle u 2 + t 2 = c to use. For example, if u(1) = 3, then c = = 10. Since u(1) < 0, we have u(t) = 10 t 2. Dan Sloughter (Furman University) Mathematics 22: Lecture 7 January 15, / 8

11 Consider the equation du dt = sin(t), u(1) = 4. 2tu Dan Sloughter (Furman University) Mathematics 22: Lecture 7 January 15, / 8

12 Consider the equation du dt = sin(t), u(1) = 4. 2tu Then sin(t) 2udu = dt. t Dan Sloughter (Furman University) Mathematics 22: Lecture 7 January 15, / 8

13 Consider the equation du dt = sin(t), u(1) = 4. 2tu Then sin(t) 2udu = dt. t Hence u 2 = t 1 sin(s) ds + 16 s Dan Sloughter (Furman University) Mathematics 22: Lecture 7 January 15, / 8

14 Homogeneous equations We all a differential equation of the form homogeneous. du ( u ) dt = F t Dan Sloughter (Furman University) Mathematics 22: Lecture 7 January 15, / 8

15 Homogeneous equations We all a differential equation of the form homogeneous. Example: The equation du ( u ) dt = F t du dt = u t u + t is homogeneous because we may write it as du dt = u t 1 u. t + 1 Dan Sloughter (Furman University) Mathematics 22: Lecture 7 January 15, / 8

16 Solving a homogeneous equation To solve du ( u ) dt = F, let y = u, that is u = ty. t t Dan Sloughter (Furman University) Mathematics 22: Lecture 7 January 15, / 8

17 Solving a homogeneous equation To solve du ( u ) dt = F, let y = u, that is u = ty. t t Then du dt = t dy + y, so dt which is a separable equation. t dy dt + y = F (y), or 1 dy F (y) y dt = 1 t, Dan Sloughter (Furman University) Mathematics 22: Lecture 7 January 15, / 8

18 For the equation du dt = u t, the substitution u = ty gives us u + t t dy dt + y = y 1 y + 1. Dan Sloughter (Furman University) Mathematics 22: Lecture 7 January 15, / 8

19 For the equation du dt = u t, the substitution u = ty gives us u + t t dy dt + y = y 1 y + 1. Hence t dy dt = y 1 y + 1 y = y2 + 1 y + 1. Dan Sloughter (Furman University) Mathematics 22: Lecture 7 January 15, / 8

20 For the equation du dt = u t, the substitution u = ty gives us u + t t dy dt + y = y 1 y + 1. Hence t dy dt = y 1 y + 1 y = y2 + 1 y + 1. Thus y y dy = t dt. Dan Sloughter (Furman University) Mathematics 22: Lecture 7 January 15, / 8

21 For the equation du dt = u t, the substitution u = ty gives us u + t t dy dt + y = y 1 y + 1. Hence t dy dt = y 1 y + 1 y = y2 + 1 y + 1. Thus y y dy = t dt. Hence 1 2 log(y 2 + 1) + tan 1 (y) = log t + k 1. Dan Sloughter (Furman University) Mathematics 22: Lecture 7 January 15, / 8

22 (cont d) That is, 2 tan 1 (y) = log(t 2 ) log(y 2 + 1) + k 2. Dan Sloughter (Furman University) Mathematics 22: Lecture 7 January 15, / 8

23 (cont d) That is, 2 tan 1 (y) = log(t 2 ) log(y 2 + 1) + k 2. Substituting y = u t, ( tan 1 u ) = 1 t 2 log(u2 + t 2 ) + c. Dan Sloughter (Furman University) Mathematics 22: Lecture 7 January 15, / 8

24 (cont d) That is, 2 tan 1 (y) = log(t 2 ) log(y 2 + 1) + k 2. Substituting y = u t, ( tan 1 u ) = 1 t 2 log(u2 + t 2 ) + c. Check in Maxima: ode2( diff(u,t)=(u-t)/(u+t), u, t) Dan Sloughter (Furman University) Mathematics 22: Lecture 7 January 15, / 8

25 (cont d) That is, 2 tan 1 (y) = log(t 2 ) log(y 2 + 1) + k 2. Substituting y = u t, ( tan 1 u ) = 1 t 2 log(u2 + t 2 ) + c. Check in Maxima: ode2( diff(u,t)=(u-t)/(u+t), u, t) Adding an initial condition in Maxima: ic1(%, t=1, u=1) Dan Sloughter (Furman University) Mathematics 22: Lecture 7 January 15, / 8

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