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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


Transcript

Ex 9.5, 15 For each of the differential equations given in Exercises 13 to 15 , find a particular solution satisfy the given condition : 𝑑𝑦/𝑑π‘₯βˆ’3𝑦 cot⁑〖π‘₯=sin⁑〖2π‘₯;𝑦=2γ€— γ€— when π‘₯= πœ‹/2 𝑑𝑦/𝑑π‘₯βˆ’3𝑦 cot⁑〖π‘₯=sin⁑2π‘₯ γ€— 𝑑𝑦/𝑑π‘₯ + (βˆ’3 cot x) y = (sin 2x) Comparing with 𝑑𝑦/𝑑π‘₯ + Py = Q P = βˆ’3 cot x & Q = sin 2x Finding Integrating factor (IF) IF = e^∫1▒𝑝𝑑π‘₯ = e^∫1β–’γ€–βˆ’3 cot⁑〖π‘₯ 𝑑π‘₯γ€— γ€— = e^(βˆ’3∫1β–’γ€–π‘π‘œπ‘‘ π‘₯ 𝑑π‘₯γ€—) = e^(βˆ’3 log⁑|sin⁑π‘₯ | ) = e^log⁑〖|sin⁑π‘₯ |^(βˆ’3) γ€— = e^log⁑〖1/|sin^3⁑π‘₯ | γ€— = e^log⁑|π‘π‘œπ‘ π‘’π‘^3 π‘₯| = 𝒄𝒐𝒔𝒆𝒄^πŸ‘ 𝒙 Solution of differential equation is y Γ— IF = ∫1▒〖𝑄.𝐼𝐹 𝑑π‘₯γ€— Putting values y Γ— cosec3 x = ∫1β–’π’”π’Šπ’β‘γ€–πŸπ’™. 𝒄𝒐𝒔𝒆𝒄^πŸ‘ 𝒙 𝒅𝒙〗 y cosec3x = ∫1β–’(2 sin⁑〖π‘₯ cos⁑π‘₯ γ€—)/sin^3⁑π‘₯ dx y cosec3x = ∫1β–’(2 cos⁑π‘₯)/sin^2⁑π‘₯ dx y cosec3x = 2∫1β–’γ€–cos⁑π‘₯/𝑠𝑖𝑛⁑π‘₯ Γ—1/sin⁑π‘₯ γ€— dx y cosec3x = 2∫1β–’γ€–cot⁑π‘₯ π‘π‘œπ‘ π‘’π‘ π‘₯γ€— dx y cosec3x = 2 (βˆ’cosec x) + C y = (βˆ’2 π‘π‘œπ‘ π‘’π‘ π‘₯)/(π‘π‘œπ‘ π‘’π‘^3 π‘₯) + 𝐢/(π‘π‘œπ‘ π‘’π‘^3 π‘₯) y = (βˆ’2)/(π‘π‘œπ‘ π‘’π‘^2 π‘₯) + 𝐢/(π‘π‘œπ‘ π‘’π‘^3 π‘₯) y = βˆ’2 sin2 x + C sin3 x Putting x = 𝝅/𝟐 , y = 2 in (2) 2 = βˆ’2 sin2 πœ‹/2 + C sin3 πœ‹/2 2 = βˆ’2 (1)2 + C(1)3 2 = βˆ’2 + C C = 2 + 2 C = 4 Put value of C in (3) y = βˆ’2 sin2 x + C sin3 x y = βˆ’2 sin2 x + 4 sin3 x y = 4 sin3 x βˆ’ 2 sin2 x

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.