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Example 21 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‘₯ π’…π’š/𝒅𝒙 = (π’™π’š πœπ¨π¬β‘γ€–(π’š/𝒙) + π’š^𝟐 𝐬𝐒𝐧⁑(π’š/𝒙) γ€—)/(π’™π’š 𝐬𝐒𝐧⁑〖(π’š/𝒙) βˆ’ 𝒙^𝟐 𝒄𝒐𝒔(π’š/𝒙)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 π’…π’š/𝒅𝒙 = 𝒙 𝒅𝒗/𝒅𝒙 + 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 + (𝒙 𝒅𝒗)/𝒅𝒙 = (𝒗 πœπ¨π¬β‘γ€–π’— + 𝒗^𝟐 𝐬𝐒𝐧⁑𝒗 γ€—)/(𝒗 𝐬𝐒𝐧⁑〖𝒗 βˆ’ πœπ¨π¬β‘π’— 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€— ) ( 𝒙 𝒅𝒗)/𝒅𝒙 = (πŸπ’— 𝒄𝒐𝒔⁑𝒗)/(𝒗 π’”π’Šπ’β‘γ€–π’— 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‘₯ (π’•π’‚π’β‘π’—βˆ’πŸ/𝒗) dv = 2 𝒅𝒙/𝒙 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’π₯𝐨𝐠⁑〖|𝒗|γ€— = 2 log |𝒙| + log 𝐂 log |sec⁑𝑣 |βˆ’log⁑〖|𝑣|γ€— = log |π‘₯^2 | + log C log |𝒔𝒆𝒄⁑𝒗/𝒗| = log 𝒙^𝟐 + log 𝐂 log |sec⁑𝑣/𝑣| = log π‘₯^2 𝑐 𝒔𝒆𝒄⁑𝒗/𝒗 = 𝒙^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

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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 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.