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Ex 9.5, 7 - Show homogeneous: {x cos (y/x) + y sin (y/x)} y dx

Ex 9.5, 7 - Chapter 9 Class 12 Differential Equations - Part 2
Ex 9.5, 7 - Chapter 9 Class 12 Differential Equations - Part 3 Ex 9.5, 7 - Chapter 9 Class 12 Differential Equations - Part 4 Ex 9.5, 7 - Chapter 9 Class 12 Differential Equations - Part 5 Ex 9.5, 7 - Chapter 9 Class 12 Differential Equations - Part 6 Ex 9.5, 7 - Chapter 9 Class 12 Differential Equations - Part 7


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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.