Question: Solve this Differential equation. change in Temperature over the change in time dT/dt. (T) = Temperature | (t) = time PLEASE SHOW ALL WORK!!!! (EVERY

Solve this Differential equation. change in Temperature over the change in time dT/dt. (T) = Temperature | (t) = time

PLEASE SHOW ALL WORK!!!! (EVERY SINGLE STEP BY STEP BY STEP SHOW ALL SOUTIONS WHEN SOLVING THE DIFFERENTIAL EQUATION.)

Solve the Differential Equation Model.

Classify the Differential Equation as a Linear or a Non-linear.

Classify the Differential Equation as Autonomous or Non-Autonomous.

Classify the Differential Equation as Homogenous or Non- Homogenous.

Classify the Differential Equation as Partial Differential Equation (PDE) or Ordinary Differential Equation (ODE).

Solve this Differential equation. change inSolve this Differential equation. change inSolve this Differential equation. change inSolve this Differential equation. change inSolve this Differential equation. change in
\fV =-COS(D3*PI()/4.42)*SIN(D3*PI()*0.5)-0.2*EXP(D3/20)/(D3*800-1) C D E F G H K T ( F ) t (min) AT At AT/At 40 0 -COS(D3*PI()/4.42)*SIN(D3*PI()*0.5)-0.2*EXP(D3/ 40 15 0 15 0 20)/(D3*800-1) 35 30 -5 15 -0.333333333 -0.842700778 dT/d 32 45 15 -0.2 -8.36915E-05 0.8 28 60 15 -0.266666667 -0.995194956 25 75 15 -0.2 -0.000250051 0.6 20 90 15 -0.333333333 -0.720 Vertical (Value) Axis Title 15 105 15 -0.333333333 -0.000840485 0.4 10o in i w ww in in w i is 12 120 15 -0.2 -0.136217337 CHANGE IN TEMPERATURE (AT) 0.2 9 135 15 -0.2 -0.003013429 5 150 15 0.266666667 0.502385184 0 0 165 15 -0.333333333 -0.011254361 50 8 180 15 0.533333333 0.910479487 -0.2 20 195 12 15 0.8 -0.043232999 -0.4 28 210 8 15 0.533333333 0.870349505 35 225 15 0.466666667 -0.169537124 0.6 38 240 | w 15 0.2 -0.425399538 40 260 N 20 0.1 Table 1: Temperature Readings over the Time of a Single Freeze-Thaw Cycle FigDependent Variable T Temperature (F) Independent Variable t Time [min] in Change in Temperature =T2-T1 At Change in Time =t2t1 Table 2: Variable Definition Changes in Temperature During a Temperature During a Freeze-Thaw Cycle Freeze-Thaw Cycle 45 1 1.5 dT/dt=-COS(t*PI()/4.42)*SIN(t*PI()*0.5)-0.2*EXP(t/20)/(t*800-1) 40 0.8 1 35 0.6 30 0.5 0.4 25 Temperature (T) CHANGE IN TEMPERATURE (AT) 0.2 O 20 150 200 250 300 0 15 150 200 250 300 -0.5 0.2 10 -1 -0.4 5 -0.6 -1.5 TIME (T) 0 50 100 150 200 250 300 (MINUTES) Time (t) (minutes) Figure 1: Change in Temperature over Time Figure 2: Temperature over Time\f

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