Variable separation - Equation given
Example 20
Misc 13 (MCQ)
Ex 9.3, 3
Ex 9.3, 8
Example 4
Ex 9.3, 7 Important
Example 5
Ex 9.3, 6
Ex 9.3, 2
Misc 4
Example 6
Ex 9.3, 5
Ex 9.3, 14
Ex 9.3, 13
Example 7 Important
Ex 9.3, 16
Ex 9.3, 12
Ex 9.3, 11 Important
Ex 9.3, 1 Important You are here
Ex 9.3, 15 Important
Misc 7 Important
Ex 9.3, 4 Important
Ex 9.3, 9 Important
Misc 6
Ex 9.3, 10 Important
Misc 12 Important
Misc 5 Important
Variable separation - Equation given
Last updated at April 16, 2024 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