Ex 9.4, 13 - Chapter 9 Class 12 Differential Equations

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Ex 9.4, 13 For each of the differential equations in Exercises from 11 to 15 , find the particular solution satisfying the given condition : [𝑥 sin^2⁡〖(𝑦/𝑥)−𝑦〗 ]𝑑𝑥+𝑥𝑑𝑦=0;𝑦=𝜋/4 When 𝑥=1 Ex 9.4, 13 For each of the differential equations in Exercises from 11 to 15 , find the particular solution satisfying the given condition : [𝑥 sin^2⁡〖(𝑦/𝑥)−𝑦〗 ]𝑑𝑥+𝑥𝑑𝑦=0;𝑦=𝜋/4 When 𝑥=1 Differential equation can be written as [𝑥 sin^2⁡〖(𝑦/𝑥)−𝑦〗 ]𝑑𝑥+𝑥𝑑𝑦=0 x dy = − [𝑥 sin^2⁡〖(𝑦/𝑥)−𝑦〗 ]dx 𝒅𝒚/𝒅𝒙 = −[𝒙/𝒙 〖𝐬𝐢𝐧〗^𝟐⁡〖𝒚/𝒙 −𝒚/𝒙〗 ] Let F(x, y) = 𝑑𝑦/𝑑𝑥 =−[𝑥/𝑥 sin^2⁡〖𝑦/𝑥 −𝑦/𝑥〗 ] Finding F(𝝀x, 𝝀y) F(𝜆x, 𝜆y) = −[sin^2⁡〖𝜆𝑦/𝜆𝑥−𝜆𝑦/𝜆𝑥〗 ] = −[sin^2⁡〖𝑦/𝑥−𝑦/𝑥〗 ] = 𝜆° F (x, y) ∴ F(x, y) is 𝑎 homogenous function of degree zero Putting y = vx Diff w.r.t. x 𝒅𝒚/𝒅𝒙 = x 𝒅𝒗/𝒅𝒙 + v Putting value of 𝑑𝑦/𝑑𝑥 and y = vx in (1) 𝑑𝑦/𝑑𝑥 = −[𝑥/𝑥 sin^2⁡〖𝑦/𝑥 −𝑦/𝑥〗 ] v + (𝒙 𝒅𝒗)/𝒅𝒙 = −[〖𝐬𝐢𝐧〗^𝟐 𝒗𝒙/𝒙−𝒗𝒙/𝒙] v + (𝑥 𝑑𝑣)/𝑑𝑥 = −[sin^2 𝑣−𝑣] (𝑥 𝑑𝑣)/𝑑𝑥 = v − sin2 v − v (𝑥 𝑑𝑣)/𝑑𝑥 = − sin2 v 𝒅𝒗/〖𝒔𝒊𝒏〗^𝟐⁡𝒗 = −𝒅𝒙/𝒙 Integrating both sides ∫1▒𝑑𝑣/(sin^2 𝑣) = −∫1▒𝑑𝑥/𝑥 ∫1▒〖𝑐𝑜𝑠𝑒𝑐^2 𝑣 𝑑𝑣〗 = −∫1▒𝑑𝑥/𝑥 − cot v = − log |𝑥|+𝑐 log |𝒙| − cot v = C Putting value of v = 𝑦/𝑥 log |𝑥| − cot (𝑦/𝑥) = C Putting x = 1 & y = 𝝅/𝟒 log 1 − cot (𝜋/4) = C 0 − 1 = C C = −1 Putting value of C in (2) log |𝑥| − cot 𝑦/𝑥 = −1 cot 𝑦/𝑥 − log |𝑥| = 1 cot 𝑦/𝑥 − log |𝑥| = log e cot 𝑦/𝑥 = log |𝑥| + log e cot 𝒚/𝒙 = log |𝒆𝒙|