Misc 14 - Chapter 9 Class 12 Differential Equations

Transcript

Misc 14 Find a particular solution of the differential equation 𝑥+1﷯ 𝑑𝑦﷮𝑑𝑥﷯=2 𝑒﷮−𝑦﷯−1 , given that 𝑦=0 when 𝑥=0 𝑥+1﷯ 𝑑𝑦﷮𝑑𝑥﷯=2 𝑒﷮−𝑦﷯−1 The variables are separable 𝑑𝑦﷮2 𝑒﷮ −𝑦 ﷯ − 1﷯ = 𝑑𝑥﷮𝑥 + 1﷯ Integrating both sides ﷮﷮ 𝑑𝑦﷮2 𝑒﷮−𝑦﷯ − 1﷯﷯ = ﷮﷮ 𝑑𝑥﷮𝑥 + 1﷯﷯ ﷮﷮ 𝑑𝑦﷮ 2﷮ 𝑒﷮𝑦﷯﷯ − 1﷯﷯ = log (x + 1) + c ﷮﷮ 𝑑𝑦﷮ 2 − 𝑒﷮𝑦﷯﷮ 𝑒﷮𝑦﷯﷯﷯﷯ = log (x + 1) + c ﷮﷮ 𝑒﷮𝑦﷯﷮2 − 𝑒﷮𝑦﷯﷯﷯ dy = log (x + 1) + C Put 2− 𝑒﷮𝑦﷯ = t 𝑒﷮𝑦﷯dy = –dt Putting value of t & dt in equation ﷮﷮ −𝑑𝑡﷮𝑡﷯﷯ = log (x + 1) + c − log t = log (x + 1) + c Putting back value of t − log (2 − 𝑒﷮𝑦﷯) = log (x + 1) + c Given y = 0 when x = 0 Putting x = 0 & y = 0 in (1) log (0 + 1) + log (2 − e0) = − C log 1 + log (2 − 1) = −C 0 + log 1 = −C 0 = −C C = 0 Putting value of C in (1) − log (2 − ey) = log (x + 1) + c −log (2 − ey) = log (x + 1) + 0 log (2 − ey) = −log (x + 1) log (2 − ey) = log 1﷮𝑥 + 1﷯﷯ 2 − ey = 1﷮𝑥 + 1﷯ ey = 2− 1﷮𝑥 + 1﷯ ey = 2𝑥 + 2 − 1﷮𝑥 + 1﷯ ey = 2𝑥 + 1﷮𝑥 + 1﷯ Taking log both sides y = log 𝟐𝒙 + 𝟏﷮𝒙+𝟏﷯﷯ , x ≠ − 1