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

Transcript

Ex 9.5, 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 𝑑𝑦/𝑑π‘₯=((π‘₯ cos⁑(𝑦/π‘₯) + 𝑦 sin⁑(𝑦/π‘₯))/(𝑦 sin⁑(𝑦/π‘₯) βˆ’ π‘₯ cos⁑〖 (𝑦/π‘₯)γ€— ))" " 𝑦/π‘₯ Step 2: Put 𝑑𝑦/𝑑π‘₯ = F (x, y) and find F(πœ†x, πœ†y) …(1) 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‘₯ 𝑑𝑦/𝑑π‘₯ = π‘₯ 𝑑𝑣/𝑑π‘₯ + 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/𝑑π‘₯ = (π‘₯ cos⁑(𝑣π‘₯/π‘₯) + 𝑣π‘₯ sin⁑(𝑣π‘₯/π‘₯))/(𝑣π‘₯ sin⁑(𝑣π‘₯/π‘₯) βˆ’ π‘₯ cos⁑(𝑣π‘₯/π‘₯) ) Γ— 𝑣π‘₯/π‘₯ 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)/𝑑π‘₯ = ( 1)/π‘₯ [(2𝑣 cos⁑𝑣)/(𝑣 𝑠𝑖𝑛 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 sin⁑〖𝑣 βˆ’cos⁑𝑣 γ€—)/(𝑣 γ€– cos〗⁑𝑣 ) dv = 2 ( 𝑑π‘₯)/π‘₯ ∫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 log⁑〖|sec⁑π‘₯/𝑣|=log⁑〖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⁑(𝑦/π‘₯)/(𝑦/π‘₯)| = log (cx2) |γ€–sec 〗⁑〖(𝑦/π‘₯) γ€—/(𝑦/π‘₯)| = cx2 |γ€–sec 〗⁑〖(𝑦/π‘₯) γ€— | = cx2 Γ— 𝑦/π‘₯ |γ€–sec 〗⁑(𝑦/π‘₯) |= C (where c1 = 1/𝑐) 1/|cos⁑(𝑦/π‘₯) | = C xy xy |π‘π‘œπ‘ (𝑦/π‘₯)| = c1 xy 𝒄𝒐𝒔|π’š/𝒙| = c1 (where c1 = 1/𝑐)

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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.