CHAPTER Let x(t) be the position (displacement) of the particle at time t. The force on the particle is given to be

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1 CHAPTER Section. Differential Equation Models.. Let y(t) be the number of bacteria at time t. The rate of change of the number of bacteria is y (t). Since this rate of change is given to be proportional to y(t), the resulting differential equation is y (t) = ky(t). Note that k is a positive constant since y (t) must be positive (i.e. the number of bacteria is growing)... Let y(t) be the number of field mice at time t. The rate of change of the number of mice is y (t). Since this rate of change is given to be inversely proportional to the square root of y(t), the resulting differential equation is y (t) = k/ y(t). Note that k is a positive constant since y (t) must be positive (i.e. the number of mice is growing)... Let y(t)be the number of ferrets at time t. The rate of change of the number of ferrets is y (t). Since this rate of change is given to be proportional to the product of y(t) and the difference between the maimum population and y(t) (i.e. 00 y(t)), the resulting differential equation is y (t) = ky(t)(00 y(t)). Note that k is a positive constant since y (t) must be positive (i.e. the number of ferrets is growing provided y(t) < 00)... Let y(t)be the quantity of radioactive substance at time t. The rate of change of the material is y (t). Since this rate of change (decay) is given to be proportional to y(t), the resulting differential equation is y (t) = ky(t). Note that k is a positive constant since y (t) must be negative (i.e. the quantity of radioactive material is decreasing)..5. Let y(t)be the quantity of material at time t. The rate of change of the material is y (t). Since this rate of change (decay) is given to be inversely proportional to y(t), the resulting differential equation is y (t) = k/y(t). Note that k is a positive constant since y (t) must be negative (i.e. the quantity of material is decreasing)..6. Let y(t) be the temperature of the potato at time t. The rate of change of the temperature is y (t). Since this rate of change is given to be proportional to the difference between the potato s temperature and that of the surrounding room (i.e. y(t) 65), the resulting differential equation is y (t) = k(y(t) 65). Note that k is a positive constant since y (t) must be negative (i.e. the potato is cooling) and since y(t) 65 > 0 (i.e. the potato is hotter than the surrounding room)..7. Let y(t) be the temperature of the thermometer at time t. The rate of change of the temperature is y (t). Since this rate of change is given to be proportional to the difference between the thermometer s temperature and that of the surrounding room (i.e. 77 y(t)), the resulting differential equation is y (t) = k(77 y(t)). Note that k is a positive constant since y (t) must be positive (i.e. the thermometer is warming) and since 77 y(t) > 0 (i.e. the thermometer is cooler than the surrounding room)..8. Let (t) be the position (displacement) of the particle at time t. The force on the particle is given to be proportional to this displacement. Therefore, the force, F, is equal to k(t) where k is a positive constant. The negative sign is present since the direction of F is opposite to that of (t). Newton s law states F = ma where m is the mass of the object and a = (t) is its acceleration. Therefore, F = ma becomes k(t) = m (t), which is the differential equation governing the motion of this particle..9. Let (t) be the position (displacement) of the particle at time t. The force on the particle is given to be proportional to the square of the particle s velocity, i.e. ( (t)). Therefore, the force, F, is equal to k( (t)) where k is a positive constant. The negative sign is present since the direction of F is opposite to that of the square of the velocity. Newton s law states F = ma where m is the mass of the object and a = (t) is its acceleration. Therefore, F = ma becomes k( (t)) = m (t), which is the differential equation governing the motion of this particle..0. Let (t)be the position (displacement) of the particle at time t. The force on the particle is given to be inversely proportional to the square of this displacement. The direction of F is opposite to that of (t). Therefore, the force, F, is equal to k/[(t) (t) ] where k is a positive constant. Note that we have written (t) (t)

2 Chapter. Introduction instead of (t) since k/[(t) (t) ] is negative when (t) is positive and k/[(t) (t) ] is positive when (t) is negative. This agrees with the desired direction of F. Newton s law states F = ma where m is the mass of the object and a = (t) is its acceleration. Therefore, F = ma becomes k (t) (t) = m (t) which is the differential equation governing the motion of this particle... Let V(t)be the voltage drop across the inductor and I(t)be the current at time t. The rate of change of the current is I (t). Since the voltage drop is proportional to the rate of change of I, we obtain the differential equation V(t)= ki (t), where k is a constant. Section. Integration.. y = t +. Integrate to obtain y = t + t + C. Interval of eistence is all reals y = t + t +. Integrate to obtain y = t + t + t + C. Interval of eistence is all reals

3 .. Integration.. y = sin t + cos t. Integrate to obtain y = ( /) cos t + (/) sin t + C. Interval of eistence is all reals y = sin t cos 5t. Integrate to obtain y = ( /) cos t (/5) sin 5t + C. Interval of eistence is all reals y = t/(+t ). Use u = +t, du = t dtand get dy = (/)du/u. Integrate to obtain y = (/) ln u+c = (/) ln( + t ) + C. Interval of eistence is all reals

4 Chapter. Introduction.6. y = t/( + t ). Let u = + t, du = t dt and get dy = (/)du/u. Integrate to obtain y = (/) ln u + C = (/) ln( + t ) + C. Interval of eistence is all reals y = t e t. Integrate by parts with u = t and dv = e t to obtain y = t e t e t t dt Integrate by parts once more and obtain Interval of eistence is all reals. y = t e t tet 9 + et 7 + C y = t cos t. Integrate by parts with u = t, dv = cos t and obtain y = t sin t + cos t 9 + C

5 .. Integration 5 Interval of eistence is all reals y = e ω sin ω. Integrate by parts with u = e ω and dv = sin ω to obtain e ω sin ωdω= e ω cos ω cos ωe ω dω. Integrate by parts again with u = e ω and dv = cos ω, to obtain e ω sin ωdω= e ω cos ω sin ωe ω sin ωe ω dω Then add the integral on the right to the integral on the left, which then becomes 5 sin ωe ω dω; divide by the 5 and obtain the answer: y = ( e ω cos ω sin ωe ω) /5 + C Interval of eistence is all reals

6 6 Chapter. Introduction.0. y = sin. Integrate by parts with u = and dv = sin to obtain y = ( /) cos +(/9) sin +C. Interval of eistence is all reals = s e s. Integrate by parts with u = s and dv = e s and obtain = s e s + e s sds. Integrate by parts again with u = s and dv = e s to obtain the answer: = s e s se s e s + C. Interval of eistence is all reals Integrating by parts, we get e u cos udu= e u sin u + sin ue u du. Integrate by parts again, with U = e u and dv = sin u and obtain e u cos udu= e u sin u e u cos u cos ue u du. Add the integral on the right to the left side; then divide by and obtain the answer: y = ( e u sin u e u cos u ) / + C.

7 .. Integration 7 Interval of eistence is all reals Use partial fractions to write r = [ u + ]. u Then integrate to obtain ( u / r = (/)(ln u ln u ) + C = ln u ) + C. There are three possible intervals of eistence, {u <0}, {0 <u<}, and { <u}. In the case when the interval of eistence is {0 <u<}, the solution are plotted here Use partial fractions to write y = [ + ]. Then integrate to obtain ( / y = (/)(ln ln ) + C = ln ) + C.

8 8 Chapter. Introduction There are three possible intervals of convergence, { <0}, {0 <<}, and { << }. Here we provide a family of solutions on the interval (0, ) y = t 6. Integrate y to obtain y = t 6t + C; the initial condition y(0) = gives = C; so y(t) = t 6t +. Interval of eistence is all reals. y y = +. Integrate to obtain y = / + + C; the initial condition y(0) = gives = C; so y(t) = / +. Interval of eistence is all reals. 0 y

9 .. Integration 9.7. (t) = te t. Integrate to obtain (t) = ( /)e t +C; the initial condition (0) = gives = ( /)+C; so C = / and (t) = ( /)e t + (/). Interval of eistence is all reals..5 y r (t) = t/( + t ). Integrate to obtain r(t) = (/) ln( + t ) + C; the initial condition, r(0) = gives = (/) ln + C or C = ; so r(t) = (/) ln( + t ) +. The interval of eistence is all reals..5 y s (r) = r cos r. Integrate by parts twice with dv being the trig - term cos r and then sin r to obtain s(r) = r sin r r cos r sin r + + C. The initial condition, s(0) = gives = C so s(r) = r sin r r cos r sin r + + The interval of eistence is all reals. y

10 0 Chapter. Introduction.0. P (t) = e t cos t. Integrate by parts twice as done in the solution to Eercise above to obtain P(t) = ( e t sin t (/)e t cos t ) + C 7 The initial condition, P(0) = gives = /7 + C or C = 8/7; so P(t) = 7 The interval of eistence is all reals. ( e t sin t (/)e t cos t ) y (t) = t. Integrate to obtain (t) = ( /)( t) / + C. The initial condition, (0) = gives = ( /)() / + C = 6/ + C or C = 9/. So (t) = ( /)( t) / + 9/. Interval of eistence is t<. 0 y u () = /( 5). Integrate to obtain u() = ln 5 +C. The initial condition, u(0) = gives = ln 5 + C or C = ln 5; so u() = ln 5 ln 5. Interval of eistence is <5. y

11 .. Integration.. y (t) = t +. Partial fractions gives t(t + ) y (t) = / t + / t + Integrating, we obtain y(t) = (/) ln t +(/) ln t + +C The initial condition, y( ) = 0gives 0 = (/) ln + (/) ln + C = (/) ln + C or C = (/) ln. So y(t) = (/) ln t +(/) ln t + (/) ln Interval of eistence is <t<0. y v (r) = r. By long division, we obtain r + v (r) = r + r + Integrating, we obtain v(r) = r r + ln r + +C The initial condition, v(0) = 0 gives 0 = ln + C = C; so Interval of eistence is r>. v(r) = r r + ln r + y

12 Chapter. Introduction.5. Let s(t) be the height of the ball at time t seconds. If g = 9.8 is the gravitational constant, then s (t) = g. Integrating we obtain, s = gt + v 0, where v 0 is a constant. The initial condition s (0) = 50 gives v 0 = 50; so s (t) = gt The velocity at t = seconds is s () = g + 50 = 0.6 meters/second. Integrating s, gives s(t) = gt / + 50t + s 0. The initial condition, s(0) = givess 0 =, so s(t) = gt / + 50t +. The height at t = seconds is s() = (9/)g + 5 = 08.9 meters..6. Let s(t) be the height of the ball at time t seconds. If g = 9.8 is the gravitational constant, then s (t) = g. Integrating we obtain, s = gt + v 0, where v 0 is a constant. The initial condition s (0) = 0 (dropped from rest) gives v 0 = 0; so s (t) = gt. The velocity at t = iss () = g = 9. meters/sec. Integrating s, gives s(t) = gt / + s 0. The initial condition s(0) = 00 gives s 0 = 00, so s(t) = gt / + 00; the height at t = seconds is s() = (9/)g + 00 = 55.9 meters..7. Let s(t) be the height of the ball at time t seconds. If g = 9.8 is the gravitational constant, then s (t) = g. Integrating we obtain, s = gt + v 0, where v 0 is a constant. The initial condition s (0) = 0 gives v 0 = 0, so s (t) = gt + 0. The maimum height occurs when the velocity reaches zero, i.e. when gt + 0 = 0, or t = 0/g =. seconds. Integrating s gives s(t) = gt / + 0t + s 0. The initial condition s(0) = 6 gives s 0 = 6, so s(t) = gt / + 0t + 6. When t =., the maimum height is s(.) = meters..8. Let s(t) be the height of the ball at time t seconds. If g = 9.8 is the gravitational constant, then s (t) = g. Integrating we obtain, s = gt +v 0, where v 0 is a constant. The initial condition s (0) = 5 gives v 0 = 5, so s (t) = gt 5. Integrating again gives s(t) = gt / 5t + s 0. The initial condition s(0) = 000 gives s 0 = 000 and so s(t) = gt / 5t =.9t 5t The ball hits the ground when s(t) = 0 which occurs at approimately t =.96 seconds.