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Ex 9.4, 8 show that the given differential equation is homogeneous and solve each of them. π‘₯ 𝑑𝑦/𝑑π‘₯βˆ’π‘¦+π‘₯𝑠𝑖𝑛(𝑦/π‘₯)=0 Step 1: Find 𝑑𝑦/𝑑π‘₯ 𝒙 π’…π’š/𝒅𝒙 = y βˆ’ x sin (π’š/𝒙) Step 2: Put 𝑑𝑦/𝑑π‘₯ = F (x, y) and find F(πœ†x, πœ†y) F(x, y) = 𝑦/π‘₯ βˆ’ sin (𝑦/π‘₯) F(πœ†x, πœ†y) = ("πœ†" 𝑦)/("πœ†" π‘₯) βˆ’ sin (("πœ†" 𝑦)/("πœ†" π‘₯)) = 𝑦/π‘₯ βˆ’ sin (𝑦/π‘₯) = F(x, y) = πœ†Β° [𝐹(π‘₯, 𝑦)] ∴ F (x, y) is a homogenous function of degree 0 . So the differential equation 𝑑𝑦/𝑑π‘₯ is homogenous Step 3: Let y = vx Solving 𝑑𝑦/𝑑π‘₯= 𝑦/π‘₯ - sin (β–ˆ(𝑦@π‘₯)) Putting y = vx Diff w.r.t.x 𝑑𝑦/𝑑π‘₯ = x 𝑑𝑣/𝑑π‘₯ + v 𝑑π‘₯/𝑑π‘₯ π’…π’š/𝒅𝒙 = x 𝒅𝒗/𝒅𝒙 + v Putting value of 𝑑𝑦/𝑑π‘₯ = (π‘₯2 + 𝑦^2)/(π‘₯2 + π‘₯𝑦) and y = vx in (1) π‘₯ 𝑑𝑦/𝑑π‘₯ = y βˆ’ x sin (𝑦/π‘₯) v + (𝒙 𝒅𝒗)/𝒅𝒙 = 𝒗𝒙/𝒙 βˆ’ sin (𝒗𝒙/𝒙) v + (π‘₯ 𝑑𝑣)/𝑑π‘₯ = 𝑣 βˆ’ sin v (π‘₯ 𝑑𝑣)/𝑑π‘₯ = v βˆ’ sin v βˆ’ v (π‘₯ 𝑑𝑣)/𝑑π‘₯ = βˆ’sin⁑𝑣 𝑑𝑣/𝑑π‘₯ = (βˆ’sin⁑𝑣)/π‘₯ 𝒅𝒗/(π’”π’Šπ’ 𝒗) = (βˆ’π’…π’™)/𝒙 Integrating both sides ∫1▒〖𝑑𝑣/(𝑠𝑖𝑛 𝑣)=∫1β–’(βˆ’π‘‘π‘₯)/π‘₯γ€— ∫1β–’γ€–π‘π‘œπ‘ π‘’π‘ 𝑣 𝑑𝑣=βˆ’βˆ«1▒𝑑π‘₯/π‘₯ γ€— log |𝒄𝒐𝒔𝒆𝒄 𝒗 βˆ’π’„π’π’•β‘π’™ |=βˆ’π’π’π’ˆβ‘|𝒙|+π’π’π’ˆβ‘π’„ log |π‘π‘œπ‘ π‘’π‘ 𝑣 βˆ’cot⁑𝑣 |+log⁑|π‘₯|=log⁑𝑐 log |π‘₯(π‘π‘œπ‘ π‘’π‘ 𝑣 βˆ’cot⁑〖𝑣)γ€— |=log⁑𝑐 x (cosec v βˆ’ cot v) = C x (1/sin⁑𝑣 βˆ’cos⁑𝑣/sin⁑𝑣 ) = C x ((1βˆ’cos⁑〖𝑣)γ€—)/sin⁑𝑣 = C x(1 βˆ’ cos v) = C sin v Putting value of v = 𝑦/π‘₯ x(πŸβˆ’π’„π’π’”(π’š/𝒙)) = C sin (π’š/𝒙)

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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 14 years. He provides courses for Maths, Science and Computer Science at Teachoo