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Solution first we isolate c g 5 g differentiate 0 5 s

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− ࠵?= ࠵?࠵?Solution:First we isolatec ="g5;g"differentiate0 ="’"5’;;s5("g5;g)"gsimplify0 =2࠵?− 2࠵?࠵?࠵?A− ࠵?+ ࠵?solving for y’y’ ="g#;g’";the slope of the orthogonal trajectories to the given curve is the negativereciprocal of y’࠵?′b= −4;A=5’";"g#;gW;W"t=5 ’";"g#;gwriting equation in general form2࠵?࠵? ࠵?࠵? +࠵?+ ࠵?࠵?࠵? = 0This is a homogeneous equation but it is also exactuvu;=uwu"=2࠵?we can solve this by the method of exact equation but it is easier to solve thisby inspectionexpanding2࠵?࠵?࠵?࠵? + ࠵?࠵?࠵? + ࠵?࠵?࠵? = 0࠵? ( ࠵?࠵?) + ࠵?࠵?࠵? = 0integrating࠵?࠵?࠵?+࠵?࠵?࠵? = 0࠵?࠵? +4k࠵?k= ࠵?3࠵?࠵?࠵? + ࠵?࠵?= ࠵?Not For Commercial Use
108Activity 2Directions:Solve the following problems completely and give two (2) points for everyitem. Your total score is 10 if you will get a perfect score. Put one (1) point if you willfinish half of the items and zero (0) if no solution at all. Answers are written on theleft side corner of the questionnaire.1. Find the isogonal trajectories of the one-parameter family of curves(࠵? + ࠵?)࠵?= 1if࠵? = ࠵?࠵?࠵?࠵?࠵?࠵? 4.Ans.࠵? + ࠵?࠵? + ࠵?࠵?ʃ࠵?࠵?(࠵?5࠵?)(࠵?࠵?#࠵?࠵?#࠵?)= ࠵?2. Find the orthogonal trajectories of the one-parameter family࠵?= ࠵?+ ࠵?.Ans.࠵?࠵? = ࠵?3. Find the orthogonal trajectories of the family of curves࠵? = ࠵?࠵?.Ans.࠵?࠵?࠵?࠵?+ ࠵?࠵?= ࠵? ࠵?࠵? ࠵?࠵?+ ࠵?࠵? = ࠵?4. Find the orthogonal trajectories of the family of curves࠵?+ ࠵?= ࠵?.Ans.࠵?࠵? ∣࠵?࠵?∣ = ࠵? ࠵?࠵? ࠵? = ࠵?࠵? ࠵?࠵?࠵?࠵?࠵? ࠵? = ±࠵?࠵?5. Find the orthogonal trajectories of the family of curves࠵?+ ࠵?= ࠵?࠵?.Ans.࠵?࠵?+ ࠵?࠵?= ࠵?࠵?Progress Indicators:(See Activity 1)Watch: Youtube1. Differential equations introduction (video) Khan Academy2. Youtube what is differential equation3. Youtube Differential equations, studying the unsolvable.1095.3 Newton’s Law of Cooling (and Heating)Based on studies, it was noted that an approximation of the temperature of abody undergoing a cooling or a heating process could be obtained by using Newton’sLaw of Cooling. The Law states that the rate of change of the body temperature isproportional to the difference of the temperature of the body and the temperature ofthe surrounding medium. It must be pointed out that several assumptions are to bemade so as to simplify the differential equations that would result. Among these isthat the constant of proportionality is the same for both heating and cooling. Anotheris that the temperature of the medium is maintained constant. To simplify thedifferential equation to be processed, the condition is that the heat released orabsorbed by the body is not large enough to cause a subsequent increase or decreasein the medium temperature. The law stated in differential equation from is:࠵?"= ࠵?࠵?࠵?࠵?࠵?࠵?࠵?࠵?࠵?࠵?࠵? ࠵? ࠵?ℎ࠵? ࠵?࠵?࠵?࠵?࠵?1= ࠵?࠵?࠵?࠵?࠵?࠵?࠵?࠵?࠵?࠵?࠵? ࠵?࠵? ࠵?ℎ࠵? ࠵?࠵?࠵?࠵?࠵?࠵?࠵?"−࠵?1= ࠵?࠵?࠵?࠵?࠵?࠵?࠵?࠵?࠵?࠵?࠵? ࠵?࠵?࠵?࠵?࠵?࠵?࠵?࠵?࠵?࠵?

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Solving Ordinary Differential Equations

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