Ex 9.4, 8 - Show homogeneous: x dy/dx - y + x sin (y/x) = 0 - Ex 9.4

part 2 - Ex 9.4, 8 - Ex 9.4 - Serial order wise - Chapter 9 Class 12 Differential Equations
part 3 - Ex 9.4, 8 - Ex 9.4 - Serial order wise - Chapter 9 Class 12 Differential Equations part 4 - Ex 9.4, 8 - Ex 9.4 - Serial order wise - Chapter 9 Class 12 Differential Equations part 5 - Ex 9.4, 8 - Ex 9.4 - Serial order wise - Chapter 9 Class 12 Differential Equations

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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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