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

Transcript

Example 27 Solve the differential equation (π‘₯ π‘‘π‘¦βˆ’π‘¦ 𝑑π‘₯)𝑦 𝑠𝑖𝑛(𝑦/π‘₯)=(𝑦 𝑑π‘₯+π‘₯ 𝑑𝑦)π‘₯ cos⁑(𝑦/π‘₯) (π‘₯ π‘‘π‘¦βˆ’π‘¦ 𝑑π‘₯)𝑦 𝑠𝑖𝑛(𝑦/π‘₯)=(𝑦 𝑑π‘₯+π‘₯ 𝑑𝑦)π‘₯ cos⁑(𝑦/π‘₯) π‘₯ 𝑦 sin⁑〖(𝑦/π‘₯)π‘‘π‘¦βˆ’π‘¦^2 𝑠𝑖𝑛(𝑦/π‘₯) γ€— 𝑑π‘₯=π‘₯𝑦 cos⁑〖(𝑦/π‘₯)𝑑π‘₯+π‘₯^2 γ€— cos⁑(𝑦/π‘₯)𝑑𝑦 [π‘₯ 𝑦 sin⁑〖(𝑦/π‘₯)βˆ’π‘₯^2 π‘π‘œπ‘ (𝑦/π‘₯) γ€— ]𝑑𝑦=[π‘₯𝑦 cos⁑〖(𝑦/π‘₯)+𝑦^2 γ€— sin⁑(𝑦/π‘₯) ]𝑑π‘₯ 𝑑𝑦/𝑑π‘₯ = (π‘₯𝑦 cos⁑〖(𝑦/π‘₯) + 𝑦^2 sin⁑(𝑦/π‘₯) γ€—)/(π‘₯𝑦 sin⁑〖(𝑦/π‘₯) βˆ’ π‘₯^2 π‘π‘œπ‘ (𝑦/π‘₯)γ€— ) Dividing numerator & denominator by x2 𝑑𝑦/𝑑π‘₯ = (𝑦/π‘₯ cos⁑〖(𝑦/π‘₯) + (𝑦/π‘₯)^2 cos⁑(𝑦/π‘₯) γ€—)/(𝑦/π‘₯ sin⁑〖(𝑦/π‘₯) βˆ’ cos⁑(𝑦/π‘₯) γ€— ) Putting y = vx. Differentiating w.r.t. x 𝑑𝑦/𝑑π‘₯ = π‘₯ 𝑑𝑣/𝑑π‘₯ + v Putting value of 𝑑𝑦/𝑑π‘₯ and y in (1) 𝑑𝑦/𝑑π‘₯ = (𝑦/π‘₯ cos⁑〖(𝑦/π‘₯) + (𝑦/π‘₯)^2 cos⁑(𝑦/π‘₯) γ€—)/(𝑦/π‘₯ sin⁑〖(𝑦/π‘₯) βˆ’ cos⁑(𝑦/π‘₯) γ€— ) v + (π‘₯ 𝑑𝑣)/𝑑π‘₯=(𝑣π‘₯/π‘₯ cos⁑〖(𝑣π‘₯/π‘₯) + (𝑣^2 π‘₯^2)/π‘₯^2 sin⁑(𝑣π‘₯/π‘₯) γ€—)/(𝑣π‘₯/π‘₯ sin⁑〖(𝑣π‘₯/π‘₯) βˆ’ π‘π‘œπ‘ (𝑣π‘₯/π‘₯)γ€— ) v + (π‘₯ 𝑑𝑣)/𝑑π‘₯ = (𝑣 cos⁑〖𝑣 + 𝑣^2 sin⁑𝑣 γ€—)/(𝑣 sin⁑〖𝑣 βˆ’ cos⁑𝑣 γ€— ) x ( 𝑑𝑣)/𝑑π‘₯ = (𝑣 cos⁑〖𝑣 + 𝑣^2 sin⁑𝑣 γ€—)/(𝑣 sin⁑〖𝑣 βˆ’ cos⁑𝑣 γ€— ) βˆ’ v x ( 𝑑𝑣)/𝑑π‘₯ = (𝑣 cos⁑〖𝑣 + 𝑣^2 sin⁑〖𝑣 βˆ’ 𝑣(𝑣 sin⁑〖𝑣 βˆ’ cos⁑〖𝑣)γ€— γ€— γ€— γ€—)/(𝑣 sin⁑〖𝑣 βˆ’ cos⁑𝑣 γ€— ) x ( 𝑑𝑣)/𝑑π‘₯ = (𝑣 cos⁑〖𝑣 + 𝑣^2 sin⁑〖𝑣 βˆ’ 𝑣^2 sin 𝑣〗 + 𝑣 cos⁑𝑣 γ€—)/(𝑣 sin⁑〖𝑣 βˆ’ cos⁑𝑣 γ€— ) ( π‘₯ 𝑑𝑣)/𝑑π‘₯ = (2𝑣 cos⁑𝑣)/(𝑣 sin⁑〖𝑣 βˆ’ cos⁑𝑣 γ€— ) ((𝑣 sin⁑〖𝑣 βˆ’ cos⁑𝑣 γ€—)/γ€–v cos〗⁑𝑣 )𝑑𝑣=2 𝑑π‘₯/π‘₯ ((𝑣 sin⁑𝑣)/γ€–v cos〗⁑𝑣 βˆ’cos⁑𝑣/γ€–v cos〗⁑𝑣 ) dv = 2 𝑑π‘₯/π‘₯ (tanβ‘π‘£βˆ’1/𝑣) dv = 2 𝑑π‘₯/π‘₯ Integrating both sides ∫1β–’γ€–(tan⁑〖𝑣 βˆ’1/𝑣〗 )𝑑𝑣=2∫1▒𝑑π‘₯/π‘₯γ€— ∫1β–’tan⁑〖𝑣 𝑑𝑣 βˆ’ γ€— ∫1▒𝑑𝑣/𝑣 = 2 ∫1▒𝑑π‘₯/π‘₯ log |sec⁑𝑣 |βˆ’log⁑〖|𝑣|γ€— = 2 log |π‘₯| + log C log |sec⁑𝑣 |βˆ’log⁑〖|𝑣|γ€— = log |π‘₯^2 | + log C log |sec⁑𝑣/𝑣| = log π‘₯^2 + log C log |sec⁑𝑣/𝑣| = log π‘₯^2 𝑐 sec⁑𝑣/𝑣 = π‘₯^2 𝑐 Putting back value of v = 𝑦/π‘₯ γ€–sec 〗⁑(𝑦/π‘₯)/((𝑦/π‘₯) ) = π‘₯^2 𝑐 sec⁑(𝑦/π‘₯) = (𝑦/π‘₯) π‘₯^2 𝑐 sec (π’š/𝒙) = C xy (As log a – log b = log π‘Ž/𝑏 & log a + log b = log ab)

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.