Last updated at Feb. 13, 2025 by Teachoo
Ex 9.3, 1 For each of the differential equations in Exercises 1 to 10, find the general solution : ππ¦/ππ₯=(1 β cosβ‘π₯)/(1 + cosβ‘π₯ ) ππ¦/ππ₯ = (1 β cosβ‘π₯)/(1 + cosβ‘π₯ ) We know that cos 2x = 2cos2 x β 1 Putting x = π₯/2 cos 2π₯/2 = 2 cos2 π₯/2 β 1 cos x = 2 cos2 π₯/2 β 1 1 + cos x = 2cos2 π₯/2 We know cos 2x = 1 β 2sin2 x Putting x = π₯/2 cos2 2π₯/2 = 1 β 2 sin2 π₯/2 cos x = 1 β 2sin2 π₯/2 1 β cos x = 2sin2 π₯/2 Putting values in equation π π/π π = (π γγπ¬π’π§γ^π γβ‘γπ/πγ )/(π γγππ¨π¬γ^π γβ‘γπ/πγ ) ππ¦ = (γsin^2 γβ‘γπ₯/2γ )/cos^(2 )β‘γπ₯/2γ ππ₯ ππ¦ = tan2 π₯/2 ππ₯ Putting tan2 π₯/2 = sec2 π₯/2 β 1 π π = ("sec2 " π/π " β 1" )π π Integrating both sides We know that tan2 x + 1 = sec2 x tan2 x = sec2 x β 1 Putting x = π₯/2 tan2 π/π = sec2 π/π β 1 β«1βπ π = β«1β(ππππ π/πβπ) π π y = β«1βγπ ππ2 π₯/2 ππ₯β β«1βππ₯γ y = π/(π/π) tan π/π β x + C y = 2 tan π₯/2 β x + c y = 2 tan π₯/2 β x + C y = π tan π/π β x + C is the general solution
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Ex 9.3, 23 (MCQ)
About the Author
Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 15 years. He provides courses for Maths, Science and Computer Science at Teachoo