Check sibling questions

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Ex 9.5, 1 In each of the Exercise 1 to 10 , show that the given differential equation is homogeneous and solve (π‘₯^2+π‘₯𝑦)𝑑𝑦=(π‘₯^2+𝑦^2 )𝑑π‘₯ (π‘₯^2+π‘₯𝑦)𝑑𝑦 =(π‘₯^2+𝑦^2 )𝑑π‘₯ Step 1: Find 𝑑𝑦/𝑑π‘₯ (x2 + xy)dy = (π‘₯^2+𝑦^2 )𝑑π‘₯ 𝑑𝑦/𝑑π‘₯ = (π‘₯^2 + 𝑦^2)/(π‘₯^2 + π‘₯𝑦) Step 2: Putting F(x , y) = 𝑑𝑦/𝑑π‘₯ and finding F(πœ†x, πœ†y) F(x, y) = (π‘₯^2 + 𝑦^2)/(π‘₯^2 + π‘₯𝑦) Finding F(𝝀x, 𝝀y) F(πœ†x, πœ†y) = ((πœ†γ€–π‘₯)γ€—^2+(πœ†γ€–π‘¦)γ€—^2)/((πœ†γ€–π‘₯)γ€—^2+πœ†π‘₯ + πœ†π‘¦) = (πœ†^2 π‘₯^2 + πœ†^2 𝑦^2)/(πœ†^2 π‘₯^2 + πœ†^2 π‘₯𝑦) = (πœ†^2 (π‘₯^2 𝑦^2))/(πœ†^2 (π‘₯^2+ π‘₯𝑦)) = (π‘₯^2 𝑦^2)/(π‘₯^2 + π‘₯𝑦) = F(x, y) So, F(πœ†x, πœ†y) = F(x, y) = πœ†0 F (x, y) Thus, F(x, y) is a homogenous equation function of order zero Therefore 𝑑𝑦/𝑑π‘₯ is a homogenous differential equation Step 3 : Solving 𝑑𝑦/𝑑π‘₯ by putting y = vx Putting y = vx Diff w.r.t.x 𝑑𝑦/𝑑π‘₯ = x𝑑𝑣/𝑑π‘₯ + v 𝑑π‘₯/𝑑π‘₯ π’…π’š/𝒅𝒙 = x 𝒅𝒗/𝒅𝒙 + v Putting value of 𝑑𝑦/𝑑π‘₯ and y = vx in (1) 𝑑𝑦/𝑑π‘₯ = (π‘₯^2 + 𝑦^2)/(π‘₯^2 + π‘₯𝑦) x 𝒅𝒗/𝒅𝒙 + v = (π’™πŸ + (𝒗𝒙)^𝟐)/(π’™πŸ + 𝒙(𝒗𝒙)) x 𝑑𝑣/𝑑π‘₯ + v = (π’™πŸ(𝟏 + 𝒗^𝟐))/(π‘₯2 + π‘₯2𝑣) x 𝑑𝑣/𝑑π‘₯ + v = (π’™πŸ(1 + 𝑣^2))/(π’™πŸ(1 + 𝑣)) x 𝑑𝑣/𝑑π‘₯ + v = (1 + 𝑣^2)/(1 + 𝑣) x 𝑑𝑣/𝑑π‘₯ = (1 + 𝑣^2)/(1 + 𝑣)βˆ’π’— x 𝑑𝑣/𝑑π‘₯ = (1 + 𝑣^2 βˆ’ 𝑣 βˆ’ 𝑣^2)/(1+𝑣) x 𝒅𝒗/𝒅𝒙 = (𝟏 βˆ’ 𝒗)/(𝟏 + 𝒗) ((1 + 𝑣))/((1 βˆ’ 𝑣)) dv = 𝑑π‘₯/π‘₯ βˆ’((𝑣 + 1)/(𝑣 βˆ’ 1)) dv = 𝑑π‘₯/π‘₯ ((𝒗 + 𝟏)/(𝒗 βˆ’ 𝟏)) dv = (βˆ’π’…π’™)/𝒙 Integrating both sides ∫1β–’((𝑣 + 1)/(𝑣 βˆ’ 1)) 𝑑𝑣=βˆ’βˆ«1▒𝑑π‘₯/π‘₯ ∫1β–’((𝒗 + 𝟏)/(𝒗 βˆ’ 𝟏)) 𝒅𝒗 = βˆ’log|𝒙|+𝒄 Let I = ∫1β–’((𝑣 + 1)/(𝑣 βˆ’ 1)) 𝑑𝑣 Solving I I = ∫1β–’((𝑣 + 1 βˆ’ 1 + 1)/(𝑣 βˆ’ 1)) 𝑑𝑣 I = ∫1β–’((𝒗 βˆ’ 𝟏 + 𝟐)/(𝒗 βˆ’ 𝟏)) 𝑑𝑣 I = ∫1β–’((𝒗 βˆ’ 𝟏 )/(𝒗 βˆ’ 𝟏)+𝟐/(𝒗 βˆ’ 𝟏)) 𝑑𝑣 I = ∫1β–’(1+2/(𝑣 βˆ’ 1)) 𝑑𝑣 I = ∫1▒𝑑𝑣+∫1β–’2/(𝑣 βˆ’ 1) 𝑑𝑣 I = 𝒗+𝟐 π₯𝐨𝐠⁑〖|π’—βˆ’πŸ|γ€— Putting v = y/x I = 𝑦/π‘₯+2 log⁑|𝑦/π‘₯βˆ’1| I = π’š/𝒙+𝟐 π’π’π’ˆβ‘|(π’š βˆ’ 𝒙)/𝒙| Putting value of I in (2) π’š/𝒙+𝟐 π’π’π’ˆβ‘|(π’š βˆ’ 𝒙)/𝒙|=βˆ’π’π’π’ˆβ‘|𝒙|+π‘ͺ 𝑦/π‘₯+π’π’π’ˆβ‘γ€–|(π’š βˆ’ 𝒙)/𝒙|^𝟐 γ€—+log⁑|π‘₯|=𝐢 𝑦/π‘₯+log⁑|(𝑦 βˆ’ π‘₯)^2/π‘₯^2 |+log⁑|π‘₯|=𝐢 𝑦/π‘₯+log⁑|(π’š βˆ’ 𝒙)^𝟐/𝒙^𝟐 Γ— 𝒙|=𝐢 𝑦/π‘₯+log⁑|(𝑦 βˆ’ π‘₯)^2/π‘₯|=𝐢 π’π’π’ˆβ‘|(π’š βˆ’ 𝒙)^𝟐/𝒙|=πΆβˆ’π‘¦/π‘₯ (π’š βˆ’ 𝒙)^𝟐/𝒙= 𝒆^(π‘ͺ βˆ’ π’š/𝒙) (𝑦 βˆ’ π‘₯)^2/π‘₯ = 𝑒^𝑐 Γ— 𝑒^(βˆ’ 𝑦/π‘₯) (𝑦 βˆ’ π‘₯)^2/π‘₯ = 𝑐 𝑒^(βˆ’ 𝑦/π‘₯) (π’™βˆ’π’š)^𝟐 = 𝒄𝒙 𝒆^(βˆ’ π’š/𝒙) (𝐴𝑠 π‘Ž log⁑𝑏=log⁑〖𝑏^π‘Ž γ€—) (𝐴𝑠 logβ‘π‘Ž+log⁑𝑏=logβ‘π‘Žπ‘)

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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.