#$&* course MTH-279 07/29 around 10:35PMJust turning in some work I did leading up to test two. Query 16 Differential Equations
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique rating: OK ********************************************* Question: Find the general solution to 8 y '' - 6 y ' + y = 0 and the unique solution for the initial conditions y (1) = 4, y ' (1) = 3/2. How does the solution behave as t -> infinity, and as t -> -infinity>? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 8x^2 - 6x + 1 = 0 x = -B +/- sqrt B^2 - 4AC / 2A x = 6 +/- sqrt 4 / 16 x = 6 +/- 2/16 x1 = 6 + 2 / 16 -> 8/16 -> x1 = ½ x2 = 6 - 2 / 16 -> x2 = ¼ use y = C1e^lambda t + C2e^lambda t 4 = C1 e ^ ½ + C2 e^1/4 - eq1 ---- y’ = ½ C1 e^ 1/2t + ¼ C2e^1/4t 3/2 = ½ C1e^1/2 + ¼ C2 e^1/4 - eq2 --- -3 = -1 C1e^1/2 - ½ C2 E^1/4 (multiply by negative to cancel) 1 = ½ C2e^1/4 2= C2 e^1/4 C2 = 2/e^1/4 ----- 4 = C1e^1/2 + 2/e^1/4 * e^1/4 4 = C1e^1/2 + 2 2=C1e^1/2 2/e^1/2 = C1 ---- y = 2/e^1/2 * e^1/2t + 2/e^1/4 * e^1/4t --- as t approaches infinity, y approaches infinity as t approaches -infinity, y approaches 0 Confidence rating: 3
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique rating: OK ********************************************* Question: Solve the equation m ( r '' - Omega^2 r) = - k r ' for r(0) = 0, r ' (0) = v_0. Omega is the angular velocity of a centrifuge, m is the mass of a particle and k is drag force constant. Physics students will recognize that m r Omega^2 is the centripetal force required to keep an object of mass m moving in a circle of radius r at angular velocity Omega. The equation models the motion of a particle at the axis which is given initial radial velocity v_0. The mass m of a particle is proportional to its volume, while the drag constant is proportional to its cross-sectional area. Assuming all particles are geometrically similar (and most likely spherical, though this is not necessary as long as they are geometrically similar, but for the sake of an accurate experiment spheres would be preferable), how then does k / m change as the diameter of particles increases? If Omega = 20 revolutions / minute and v_0 = 1 cm / second, with k / m = 4 s^-1, find r( 2 seconds ). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Confidence rating:
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating:
#$&* course MTH-279 07/29 around 10:35PMJust turning in some work I did leading up to test two. Query 16 Differential Equations
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique rating: OK ********************************************* Question: Find the general solution to 8 y '' - 6 y ' + y = 0 and the unique solution for the initial conditions y (1) = 4, y ' (1) = 3/2. How does the solution behave as t -> infinity, and as t -> -infinity>? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 8x^2 - 6x + 1 = 0 x = -B +/- sqrt B^2 - 4AC / 2A x = 6 +/- sqrt 4 / 16 x = 6 +/- 2/16 x1 = 6 + 2 / 16 -> 8/16 -> x1 = ½ x2 = 6 - 2 / 16 -> x2 = ¼ use y = C1e^lambda t + C2e^lambda t 4 = C1 e ^ ½ + C2 e^1/4 - eq1 ---- y’ = ½ C1 e^ 1/2t + ¼ C2e^1/4t 3/2 = ½ C1e^1/2 + ¼ C2 e^1/4 - eq2 --- -3 = -1 C1e^1/2 - ½ C2 E^1/4 (multiply by negative to cancel) 1 = ½ C2e^1/4 2= C2 e^1/4 C2 = 2/e^1/4 ----- 4 = C1e^1/2 + 2/e^1/4 * e^1/4 4 = C1e^1/2 + 2 2=C1e^1/2 2/e^1/2 = C1 ---- y = 2/e^1/2 * e^1/2t + 2/e^1/4 * e^1/4t --- as t approaches infinity, y approaches infinity as t approaches -infinity, y approaches 0 Confidence rating: 3
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique rating: OK ********************************************* Question: Solve the equation m ( r '' - Omega^2 r) = - k r ' for r(0) = 0, r ' (0) = v_0. Omega is the angular velocity of a centrifuge, m is the mass of a particle and k is drag force constant. Physics students will recognize that m r Omega^2 is the centripetal force required to keep an object of mass m moving in a circle of radius r at angular velocity Omega. The equation models the motion of a particle at the axis which is given initial radial velocity v_0. The mass m of a particle is proportional to its volume, while the drag constant is proportional to its cross-sectional area. Assuming all particles are geometrically similar (and most likely spherical, though this is not necessary as long as they are geometrically similar, but for the sake of an accurate experiment spheres would be preferable), how then does k / m change as the diameter of particles increases? If Omega = 20 revolutions / minute and v_0 = 1 cm / second, with k / m = 4 s^-1, find r( 2 seconds ). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Confidence rating:
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!
#$&* course MTH-279 07/29 around 10:35PMJust turning in some work I did leading up to test two. Query 16 Differential Equations
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique rating: OK ********************************************* Question: Find the general solution to 8 y '' - 6 y ' + y = 0 and the unique solution for the initial conditions y (1) = 4, y ' (1) = 3/2. How does the solution behave as t -> infinity, and as t -> -infinity>? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 8x^2 - 6x + 1 = 0 x = -B +/- sqrt B^2 - 4AC / 2A x = 6 +/- sqrt 4 / 16 x = 6 +/- 2/16 x1 = 6 + 2 / 16 -> 8/16 -> x1 = ½ x2 = 6 - 2 / 16 -> x2 = ¼ use y = C1e^lambda t + C2e^lambda t 4 = C1 e ^ ½ + C2 e^1/4 - eq1 ---- y’ = ½ C1 e^ 1/2t + ¼ C2e^1/4t 3/2 = ½ C1e^1/2 + ¼ C2 e^1/4 - eq2 --- -3 = -1 C1e^1/2 - ½ C2 E^1/4 (multiply by negative to cancel) 1 = ½ C2e^1/4 2= C2 e^1/4 C2 = 2/e^1/4 ----- 4 = C1e^1/2 + 2/e^1/4 * e^1/4 4 = C1e^1/2 + 2 2=C1e^1/2 2/e^1/2 = C1 ---- y = 2/e^1/2 * e^1/2t + 2/e^1/4 * e^1/4t --- as t approaches infinity, y approaches infinity as t approaches -infinity, y approaches 0 Confidence rating: 3
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique rating: OK ********************************************* Question: Solve the equation m ( r '' - Omega^2 r) = - k r ' for r(0) = 0, r ' (0) = v_0. Omega is the angular velocity of a centrifuge, m is the mass of a particle and k is drag force constant. Physics students will recognize that m r Omega^2 is the centripetal force required to keep an object of mass m moving in a circle of radius r at angular velocity Omega. The equation models the motion of a particle at the axis which is given initial radial velocity v_0. The mass m of a particle is proportional to its volume, while the drag constant is proportional to its cross-sectional area. Assuming all particles are geometrically similar (and most likely spherical, though this is not necessary as long as they are geometrically similar, but for the sake of an accurate experiment spheres would be preferable), how then does k / m change as the diameter of particles increases? If Omega = 20 revolutions / minute and v_0 = 1 cm / second, with k / m = 4 s^-1, find r( 2 seconds ). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Confidence rating:
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!#*&!
#$&* course MTH-279 07/29 around 10:35PMJust turning in some work I did leading up to test two. Query 16 Differential Equations
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique rating: OK ********************************************* Question: Find the general solution to 8 y '' - 6 y ' + y = 0 and the unique solution for the initial conditions y (1) = 4, y ' (1) = 3/2. How does the solution behave as t -> infinity, and as t -> -infinity>? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 8x^2 - 6x + 1 = 0 x = -B +/- sqrt B^2 - 4AC / 2A x = 6 +/- sqrt 4 / 16 x = 6 +/- 2/16 x1 = 6 + 2 / 16 -> 8/16 -> x1 = ½ x2 = 6 - 2 / 16 -> x2 = ¼ use y = C1e^lambda t + C2e^lambda t 4 = C1 e ^ ½ + C2 e^1/4 - eq1 ---- y’ = ½ C1 e^ 1/2t + ¼ C2e^1/4t 3/2 = ½ C1e^1/2 + ¼ C2 e^1/4 - eq2 --- -3 = -1 C1e^1/2 - ½ C2 E^1/4 (multiply by negative to cancel) 1 = ½ C2e^1/4 2= C2 e^1/4 C2 = 2/e^1/4 ----- 4 = C1e^1/2 + 2/e^1/4 * e^1/4 4 = C1e^1/2 + 2 2=C1e^1/2 2/e^1/2 = C1 ---- y = 2/e^1/2 * e^1/2t + 2/e^1/4 * e^1/4t --- as t approaches infinity, y approaches infinity as t approaches -infinity, y approaches 0 Confidence rating: 3
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique rating: OK ********************************************* Question: Solve the equation m ( r '' - Omega^2 r) = - k r ' for r(0) = 0, r ' (0) = v_0. Omega is the angular velocity of a centrifuge, m is the mass of a particle and k is drag force constant. Physics students will recognize that m r Omega^2 is the centripetal force required to keep an object of mass m moving in a circle of radius r at angular velocity Omega. The equation models the motion of a particle at the axis which is given initial radial velocity v_0. The mass m of a particle is proportional to its volume, while the drag constant is proportional to its cross-sectional area. Assuming all particles are geometrically similar (and most likely spherical, though this is not necessary as long as they are geometrically similar, but for the sake of an accurate experiment spheres would be preferable), how then does k / m change as the diameter of particles increases? If Omega = 20 revolutions / minute and v_0 = 1 cm / second, with k / m = 4 s^-1, find r( 2 seconds ). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Confidence rating:
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!#*&!#*&!
#$&* course MTH-279 07/29 around 10:35PMJust turning in some work I did leading up to test two. Query 16 Differential Equations
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique rating: OK ********************************************* Question: Find the general solution to 8 y '' - 6 y ' + y = 0 and the unique solution for the initial conditions y (1) = 4, y ' (1) = 3/2. How does the solution behave as t -> infinity, and as t -> -infinity>? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 8x^2 - 6x + 1 = 0 x = -B +/- sqrt B^2 - 4AC / 2A x = 6 +/- sqrt 4 / 16 x = 6 +/- 2/16 x1 = 6 + 2 / 16 -> 8/16 -> x1 = ½ x2 = 6 - 2 / 16 -> x2 = ¼ use y = C1e^lambda t + C2e^lambda t 4 = C1 e ^ ½ + C2 e^1/4 - eq1 ---- y’ = ½ C1 e^ 1/2t + ¼ C2e^1/4t 3/2 = ½ C1e^1/2 + ¼ C2 e^1/4 - eq2 --- -3 = -1 C1e^1/2 - ½ C2 E^1/4 (multiply by negative to cancel) 1 = ½ C2e^1/4 2= C2 e^1/4 C2 = 2/e^1/4 ----- 4 = C1e^1/2 + 2/e^1/4 * e^1/4 4 = C1e^1/2 + 2 2=C1e^1/2 2/e^1/2 = C1 ---- y = 2/e^1/2 * e^1/2t + 2/e^1/4 * e^1/4t --- as t approaches infinity, y approaches infinity as t approaches -infinity, y approaches 0 Confidence rating: 3
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique rating: OK ********************************************* Question: Solve the equation m ( r '' - Omega^2 r) = - k r ' for r(0) = 0, r ' (0) = v_0. Omega is the angular velocity of a centrifuge, m is the mass of a particle and k is drag force constant. Physics students will recognize that m r Omega^2 is the centripetal force required to keep an object of mass m moving in a circle of radius r at angular velocity Omega. The equation models the motion of a particle at the axis which is given initial radial velocity v_0. The mass m of a particle is proportional to its volume, while the drag constant is proportional to its cross-sectional area. Assuming all particles are geometrically similar (and most likely spherical, though this is not necessary as long as they are geometrically similar, but for the sake of an accurate experiment spheres would be preferable), how then does k / m change as the diameter of particles increases? If Omega = 20 revolutions / minute and v_0 = 1 cm / second, with k / m = 4 s^-1, find r( 2 seconds ). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Confidence rating:
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!#*&!#*&!#*&!