Check sibling questions

Slide114.JPG

Slide115.JPG
Slide116.JPG
Slide117.JPG
Slide118.JPG
Slide119.JPG
Slide120.JPG

Now learn Economics at Teachoo for Class 12


Transcript

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‘)

Davneet Singh's photo - Teacher, Engineer, Marketer

Made by

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.