Slide1.JPG

Slide2.JPG
Slide3.JPG
Slide4.JPG
Slide5.JPG Slide6.JPG Slide7.JPG


Transcript

Ex 9.4, 1 In each of the Exercise 1 to 10 , show that the given differential equation is homogeneous and solve (x^2+xy)dy=(x^2+y^2 )𝑑x (π‘₯^2+π‘₯𝑦)𝑑𝑦 =(π‘₯^2+𝑦^2 )𝑑π‘₯ Step 1: Find 𝑑𝑦/𝑑π‘₯ (x2 + xy)dy = (π‘₯^2+𝑦^2 )𝑑π‘₯ π’…π’š/𝒅𝒙 = (𝒙^𝟐 + π’š^𝟐)/(𝒙^𝟐 + π’™π’š) 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‘)

Ask a doubt
Davneet Singh's photo - Co-founder, Teachoo

Made by

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.