Ex 9.4, 5 - Class 12

Transcript

Ex 9.4, 5 For each of the differential equations in Exercises 1 to 10, find the general solution : 𝑒﷮𝑥﷯+ 𝑒﷮−𝑥﷯﷯𝑑𝑦− 𝑒﷮𝑥﷯− 𝑒﷮−𝑥﷯﷯𝑑𝑥=0 𝑒﷮𝑥﷯+ 𝑒﷮−𝑥﷯﷯𝑑𝑦− 𝑒﷮𝑥﷯− 𝑒﷮−𝑥﷯﷯𝑑𝑥=0 𝑒﷮𝑥﷯+ 𝑒﷮−𝑥﷯﷯ dy = 𝑒﷮𝑥﷯− 𝑒﷮−𝑥﷯﷯ dx 𝑑𝑦﷮𝑑𝑥﷯ = 𝑒﷮𝑥﷯ − 𝑒﷮−𝑥﷯﷮ 𝑒﷮𝑥﷯ + 𝑒﷮−𝑥﷯﷯ dx 𝑑𝑦 = 𝑒﷮𝑥﷯ − 𝑒﷮−𝑥﷯﷮ 𝑒﷮𝑥﷯ + 𝑒﷮−𝑥﷯﷯ dx Integrating both sides. ﷮﷮𝑑𝑦﷯ = ﷮﷮ 𝑒﷮𝑥﷯ − 𝑒﷮−𝑥﷯﷮ 𝑒﷮𝑥﷯ + 𝑒﷮−𝑥﷯﷯﷯ dx 𝑦 = ﷮﷮ 𝑒﷮𝑥﷯ − 𝑒﷮−𝑥﷯﷮ 𝑒﷮𝑥﷯ + 𝑒﷮−𝑥﷯﷯﷯ dx Let t = 𝑒﷮𝑥﷯+ 𝑒﷮−𝑥﷯ 𝑑𝑡﷮𝑑𝑥﷯ = 𝑒﷮𝑥﷯− 𝑒﷮−𝑥﷯﷯ dx dx = 𝑑𝑡﷮ 𝑒﷮𝑥﷯ − 𝑒﷮−𝑥﷯﷯ Substituting values in (1), we get ﷮﷮𝑑𝑦﷯ = ﷮﷮ 𝑒﷮𝑥 ﷯− 𝑒﷮−𝑥﷯﷮𝑡﷯﷯ 𝑑𝑡﷮ 𝑒﷮𝑥﷯ − 𝑒﷮−𝑥﷯﷯ . ﷮﷮𝑑𝑦﷯ = ﷮﷮ 𝑑𝑡﷮𝑡﷯ ﷯ y = log 𝑡﷯+𝑐 Putting back t = 𝑒﷮𝑥﷯− 𝑒﷮−𝑥﷯ y = log 𝑒﷮𝑥﷯− 𝑒﷮−𝑥﷯﷯ + C y = log ( 𝒆﷮𝒙﷯− 𝒆﷮−𝒙﷯) + C