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Ex 9.4, 7 Show that the given differential equation is homogeneous and solve each of them. {π‘₯π‘π‘œπ‘ (𝑦/π‘₯)+𝑦 sin⁑(𝑦/π‘₯) }𝑦 𝑑π‘₯={𝑦𝑠𝑖𝑛(𝑦/π‘₯)βˆ’π‘₯ cos⁑(𝑦/π‘₯) }π‘₯ 𝑑𝑦 Step 1: Find 𝑑𝑦/𝑑π‘₯ {π‘₯π‘π‘œπ‘ (𝑦/π‘₯)+𝑦 sin⁑(𝑦/π‘₯) }𝑦 𝑑π‘₯={𝑦𝑠𝑖𝑛(𝑦/π‘₯)βˆ’π‘₯ cos⁑(𝑦/π‘₯) }π‘₯ 𝑑𝑦 π’…π’š/𝒅𝒙=((𝒙 𝒄𝒐𝒔⁑(π’š/𝒙) + π’š π’”π’Šπ’β‘(π’š/𝒙))/(π’š π’”π’Šπ’β‘(π’š/𝒙) βˆ’ 𝒙 𝒄𝒐𝒔⁑〖 (π’š/𝒙)γ€— ))" " π’š/𝒙 Step 2: Put 𝑑𝑦/𝑑π‘₯ = F (x, y) and find F(πœ†x, πœ†y) F(x, y) = ((π‘₯ cos⁑(𝑦/π‘₯) + 𝑦 sin⁑(𝑦/π‘₯))/(𝑦 sin⁑(𝑦/π‘₯) βˆ’ π‘₯ cos⁑〖 (𝑦/π‘₯)γ€— )) 𝑦/π‘₯ F(πœ†x, πœ†y) = ((πœ†π‘₯ cos⁑( πœ†π‘¦/πœ†π‘₯) + πœ†π‘¦ sin⁑( πœ†π‘¦/πœ†π‘₯ ))/(πœ†π‘₯ sin⁑( πœ†π‘¦/πœ†π‘₯) βˆ’ πœ†π‘₯ cos⁑〖 ( πœ†π‘¦/πœ†π‘₯ )γ€— )) πœ†π‘¦/πœ†π‘₯ = ((π‘₯ cos⁑(𝑦/π‘₯) + 𝑦 sin⁑(𝑦/π‘₯))/(𝑦 sin⁑(𝑦/π‘₯) βˆ’ π‘₯ cos⁑〖 (𝑦/π‘₯)γ€— ))" " 𝑦/π‘₯ = F(x, y) ∴ F(πœ†x, πœ†y) = F(x, y) = πœ†Β° F(x, y) Hence, F(x, y) is a homogenous Function of with degree zero So, 𝑑𝑦/𝑑π‘₯ is a homogenous differential equation. Step 3: Solving 𝑑𝑦/𝑑π‘₯ by putting y = vx Putting y = vx. Differentiating w.r.t.x 𝑑𝑦/𝑑π‘₯ = x 𝑑𝑣/𝑑π‘₯+𝑣𝑑π‘₯/𝑑π‘₯ π’…π’š/𝒅𝒙 = 𝒙 𝒅𝒗/𝒅𝒙 + v Putting value of 𝑑𝑦/𝑑π‘₯ and y = vx in (1) 𝑑𝑦/𝑑π‘₯=((π‘₯ cos⁑(𝑦/π‘₯) + 𝑦 sin⁑(𝑦/π‘₯))/(𝑦 sin⁑(𝑦/π‘₯) βˆ’ π‘₯ cos⁑〖 (𝑦/π‘₯)γ€— ))" " 𝑦/π‘₯ v + x 𝒅𝒗/𝒅𝒙 = (𝒙 𝒄𝒐𝒔⁑(𝒗𝒙/𝒙) + 𝒗𝒙 π’”π’Šπ’β‘(𝒗𝒙/𝒙))/(𝒗𝒙 π’”π’Šπ’β‘(𝒗𝒙/𝒙) βˆ’ 𝒙 𝒄𝒐𝒔⁑(𝒗𝒙/𝒙) ) Γ— 𝒗𝒙/𝒙 v + x 𝑑𝑣/𝑑π‘₯ = [π‘₯π‘π‘œπ‘  𝑣 + 𝑣π‘₯ 𝑠𝑖𝑛 𝑣]𝑣/(𝑣π‘₯ sin⁑〖𝑣 βˆ’ π‘₯ cos⁑𝑣 γ€— ) v + x 𝑑𝑣/𝑑π‘₯ = π‘₯[π‘π‘œπ‘  𝑣 + 𝑣 𝑠𝑖𝑛 𝑣]𝑣/π‘₯[𝑣 𝑠𝑖𝑛 𝑣 βˆ’γ€– cos〗⁑𝑣 ] v + x 𝑑𝑣/𝑑π‘₯ = 𝑣[π‘π‘œπ‘  𝑣 + 𝑣 𝑠𝑖𝑛 𝑣]/(𝑣 𝑠𝑖𝑛 𝑣 βˆ’γ€– cos〗⁑𝑣 ) x ( 𝑑𝑣)/𝑑π‘₯ = (𝑣 cos⁑𝑣 + 𝑣^2 sin⁑𝑣)/(𝑣 𝑠𝑖𝑛 𝑣 βˆ’γ€– cos〗⁑𝑣 ) βˆ’ v x ( 𝑑𝑣)/𝑑π‘₯ = (𝑣 cos⁑𝑣 + 𝑣^2 sin⁑𝑣 βˆ’ 𝑣(𝑣 sin⁑〖𝑣 γ€—βˆ’ cos⁑𝑣 ))/(𝑣 𝑠𝑖𝑛 𝑣 βˆ’γ€– cos〗⁑𝑣 ) x ( 𝑑𝑣)/𝑑π‘₯ = (𝑣 cos⁑𝑣 + 𝑣^2 sin⁑𝑣 βˆ’ 𝑣^2 sin⁑〖𝑣 + 𝑣 cos⁑𝑣 γ€—)/(𝑣 𝑠𝑖𝑛 𝑣 βˆ’γ€– cos〗⁑𝑣 ) ( π‘₯𝑑𝑣)/𝑑π‘₯ = (2𝑣 cos⁑𝑣)/(𝑣 𝑠𝑖𝑛 𝑣 βˆ’γ€– cos〗⁑𝑣 ) ( 𝒅𝒗)/𝒅𝒙 = ( 𝟏)/𝒙 [(πŸπ’— 𝒄𝒐𝒔⁑𝒗)/(𝒗 π’”π’Šπ’ 𝒗 βˆ’γ€– 𝒄𝒐𝒔〗⁑𝒗 )] (𝑣 sin⁑〖𝑣 βˆ’ cos⁑𝑣 γ€—)/(𝑣 γ€– cos〗⁑𝑣 ) dv = 2 ( 𝑑π‘₯)/π‘₯ Integrating both sides. ∫1β–’(𝒗 π’”π’Šπ’β‘γ€–π’— βˆ’π’„π’π’”β‘π’— γ€—)/(𝒗 γ€– 𝒄𝒐𝒔〗⁑𝒗 ) dv = 2 ( 𝒅𝒙)/𝒙 ∫1β–’(𝑣 sin⁑〖𝑣 γ€—)/(𝑣 γ€– cos〗⁑𝑣 ) dv βˆ’ ∫1β–’cos⁑𝑣/(𝑣 γ€– cos〗⁑𝑣 ) dv = 2∫1▒𝑑π‘₯/π‘₯ ∫1β–’sin⁑〖𝑣 γ€—/cos⁑𝑣 dv βˆ’ ∫1▒𝑑𝑣/𝑣 dv = 2∫1▒𝑑π‘₯/π‘₯ ∫1β–’tan⁑𝑣 dv βˆ’ ∫1▒𝑑𝑣/𝑣 dv = 2∫1▒𝑑π‘₯/π‘₯ log⁑〖|sec⁑π‘₯ |βˆ’log⁑|𝑣|=2 log⁑|π‘₯| γ€— + C π’π’π’ˆβ‘γ€–|𝒔𝒆𝒄⁑𝒙/𝒗|=π’π’π’ˆβ‘γ€–π’™^𝟐 γ€—+𝒄 γ€— log⁑〖|sec⁑π‘₯/𝑣|=log⁑〖π‘₯^2 γ€—+log⁑𝑐 γ€— log⁑〖|sec⁑π‘₯/𝑣|=log⁑〖𝑐π‘₯^2 γ€— γ€— Put v = 𝑦/π‘₯ log |𝐬𝐞𝐜⁑(π’š/𝒙)/(π’š/𝒙)| = log (cx2) |γ€–sec 〗⁑〖(𝑦/π‘₯) γ€—/(𝑦/π‘₯)| = cx2 |γ€–sec 〗⁑〖(𝑦/π‘₯) γ€— | = cx2 Γ— 𝑦/π‘₯ |〖𝒔𝒆𝒄 〗⁑(π’š/𝒙) |= C 1/|cos⁑(𝑦/π‘₯) | = C xy xy |π‘π‘œπ‘ (𝑦/π‘₯)| = c1 xy 𝒄𝒐𝒔|π’š/𝒙| = c1

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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.