Misc 9 - Find particular solution: (1 + e2x)dy + (1 + y2) ex - Variable separation - Equation given

Slide10.JPG
Slide11.JPG

  1. Chapter 9 Class 12 Differential Equations
  2. Serial order wise
Ask Download

Transcript

Misc 9 Find the particular solution of the differential equation (1 + 𝑒﷮2𝑥﷯) dy + (1 + 𝑦﷮2﷯) ex dx = 0, given that y = 1 when x = 0. (1 + e2x) dy + (1 + y2) 𝑒﷮𝑥﷯ dx = 0 (1 + e2x) dy = −(1 + y2) 𝑒﷮𝑥﷯ dx 𝑑𝑦﷮𝑑𝑥﷯ = − 1 + 𝑦﷮2﷯﷯. 𝑒﷮𝑥﷯﷮1 + 𝑒2𝑥﷯ 𝑑𝑦﷮ 1 + 𝑦﷯﷮2﷯﷯ = − 𝑒﷮𝑥﷯ 𝑑𝑥﷮1 + 𝑒2𝑥﷯ Integrating both sides. ﷮﷮ 𝑑𝑦﷮ 1 + 𝑦﷮2﷯﷯﷯ = ﷮﷮ 𝑒𝑥 𝑑𝑥﷮ 1 + 𝑒﷮2𝑥﷯﷯﷯ Let t = ex Diff w.r.t.x 𝑑𝑡﷮𝑑𝑥﷯= 𝑒﷮𝑥﷯ 𝑑𝑡﷮𝑒𝑥﷯= 𝑑𝑥 ∴ Our equation becomes ﷮﷮ 𝑑𝑦﷮ 1 + 𝑦﷮2﷯﷯﷯ = − ﷮﷮ 𝑒𝑥 ﷮1 + 𝑡﷮2﷯﷯ 𝑑𝑡 ﷮𝑒𝑥 ﷯﷯ ﷮﷮ 𝑑𝑦﷮ 1 + 𝑦﷮2﷯﷯﷯ = − ﷮﷮ 𝑑𝑡 ﷮1 + 𝑡﷮2﷯﷯ ﷯ tan﷮−1﷯y = −tan﷮−1﷯t+C Putting value of t tan﷮−1﷯y = – tan﷮−1﷯ ex + c Given that y = 1 when x = 0 Put y = 1 and x = 0 in (2) tan−1 (1) = – tan−1 e° + C 𝜋﷮4﷯=− tan﷮−1﷯﷮1+𝑐﷯ 𝜋﷮4﷯= −𝜋﷮4﷯ + C 𝜋﷮4﷯ + 𝜋﷮4﷯ = C C = 𝜋﷮2﷯. Putting value of C in (1) tan−1 y = − tan−1 ex − 𝜋﷮2﷯ tan−1 y + tan−1 ex = 𝝅﷮𝟐﷯ is The required particular solution.

About the Author

Davneet Singh's photo - Teacher, Computer Engineer, Marketer
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 7 years. He provides courses for Mathematics and Science from Class 6 to 12. You can learn personally from here https://www.teachoo.com/premium/maths-and-science-classes/.
Jail