# Ex 9.4, 11 - Chapter 9 Class 12 Differential Equations

Last updated at April 8, 2024 by Teachoo

Last updated at April 8, 2024 by Teachoo

Ex 9.4, 11 For each of the differential equations in Exercises from 11 to 15 , find the particular solution satisfying the given condition : (𝑥+𝑦)𝑑𝑦+(𝑥−𝑦)𝑑𝑥=0;𝑦=1 When 𝑥=1 The differential equation can be written as (𝑥+𝑦)𝑑𝑦+(𝑥−𝑦)𝑑𝑥=0 𝑑𝑦/𝑑𝑥 = (−(𝑥 − 𝑦))/(𝑥 + 𝑦) Let F(x, y) = 𝑑𝑦/𝑑𝑥 = (−(𝑥 − 𝑦))/(𝑥 + 𝑦) Finding F(𝝀x, 𝝀y) F(𝜆x, 𝜆y) = (−(𝜆𝑥 − 𝜆𝑦))/(𝜆𝑥 + 𝜆𝑦) = (−𝜆(𝑥 − 𝑦))/(𝜆(𝑥 + 𝑦)) = (−(𝑥 − 𝑦))/(𝑥 + 𝑦) = 𝜆° F(x, y) ∴ F(x, y) is a homogenous function of degree zero Putting y = vx. Differentiating w.r.t. x 𝒅𝒚/𝒅𝒙 = 𝒙 𝒅𝒗/𝒅𝒙 + v Putting value of 𝑑𝑦/𝑑𝑥 and y = vx in (1) 𝑑𝑦/𝑑𝑥 = (−(𝑥 − 𝑦))/(𝑥 + 𝑦) 𝒗 + (𝒙 𝒅𝒗)/𝒅𝒙 = (−(𝒙 − 𝒗𝒙))/(𝒙 + 𝒗𝒙) 𝑣 + (𝑥 𝑑𝑣)/𝑑𝑥 = (−𝑥(1 − 𝑣))/(𝑥(1 + 𝑣)) 𝑣 + (𝑥 𝑑𝑣)/𝑑𝑥 = (𝑣 − 1)/(1 + 𝑣) (𝑥 𝑑𝑣)/𝑑𝑥 = (𝑣 − 1)/(1 + 𝑣)−𝑣 (𝑥 𝑑𝑣)/𝑑𝑥 = (𝑣 − 1 − 𝑣(1 + 𝑣) )/(1 + 𝑣) (𝑥 𝑑𝑣)/𝑑𝑥 = (𝑣 − 1 − 𝑣 − 𝑣^2)/(1 + 𝑣) (𝑥 𝑑𝑣)/𝑑𝑥 = (−(1 + 𝑣^2 ))/(1 + 𝑣) (𝟏 + 𝒗)/(𝟏 + 𝒗^𝟐 ) 𝒅𝒗 = (−𝒅𝒙)/𝒙 Integrating both sides ∫1▒〖(1 + 𝑣)/(1 + 𝑣^2 ) 𝑑𝑣=−∫1▒𝑑𝑥/𝑥〗 ∫1▒〖1/(𝑣^2 + 1) 𝑑𝑣+∫1▒𝑣/(𝑣^2 + 1) 𝑑𝑣=−log|𝑥|+𝐶〗 tan−1 v + ∫1▒𝒗/(𝒗^𝟐 + 𝟏) 𝒅𝒗=−𝒍𝒐𝒈|𝒙|+𝑪 Putting v2 + 1 = t 2v dv = dt v dv = 𝑑𝑡/2 Thus, our equation becomes tan−1 v + ∫1▒〖1/𝑡 × 𝑑𝑡/2 " =" −log|𝑥|+𝐶" " 〗 tan−1 v + 1/2×log〖|𝑡|〗 " ="−log|𝑥|+𝐶 Putting back value of t tan−1 v + 1/2×log〖|𝑣^2+1|〗 " ="−log|𝑥|+𝐶 Putting back value of v = 𝑦/𝑥 tan−1 𝑦/𝑥 + 1/2×log|(𝑦/𝑥)^2+1| " ="−log|𝑥|+𝐶 tan−1 𝑦/𝑥 + 1/2×log|(𝑦/𝑥)^2+1| " "+log|𝑥|=𝐶 tan−1 𝑦/𝑥 + 1/2×log|(𝑦/𝑥)^2+1| " "+2/2 log|𝑥|=𝐶 tan−1 𝑦/𝑥 + 1/2×(log|(𝑦/𝑥)^2+1| " " +2 log|𝑥| )=𝐶 tan−1 𝑦/𝑥 + 1/2×(log|𝑦^2/𝑥^2 +1| " " +log|𝑥^2 | )=𝐶 tan−1 𝑦/𝑥 + 1/2×(log((𝑦^2 + 𝑥^2)/𝑥^2 )" " +log〖𝑥^2 〗 )=𝐶 tan−1 𝑦/𝑥 + 1/2×(log((𝑦^2 + 𝑥^2)/𝑥^2 ×𝑥^2 ) )=𝐶 tan−1 𝒚/𝒙 + 𝟏/𝟐×(𝒍𝒐𝒈(𝒚^𝟐+𝒙^𝟐 ) )=𝑪 2tan−1 𝑦/𝑥 + (log(𝑦^2+𝑥^2 ) )=2𝐶 2tan−1 𝑦/𝑥 + (log(𝑦^2+𝑥^2 ) )=𝐾 Now, Putting x = 1 & y = 1 in equation 2tan−1 𝑦/𝑥 + log(𝑦^2+𝑥^2 )=𝐾 2tan−1 (1/1) + log (1^2+1^2 )= K 2tan−1 1 + log 2 = K 2 × 𝜋/4 + log 2 = K 𝜋/2 + log 2 = K K = 𝝅/𝟐 + log 2 Put value of K in (2) 2tan−1 𝑦/𝑥 + log(𝑦^2+𝑥^2 )=𝐾 2tan−1 𝑦/𝑥 + log(𝑦^2+𝑥^2 )=𝜋/2 " + log 2" 𝒍𝒐𝒈(𝒚^𝟐+𝒙^𝟐 )+ 2tan−1 𝒚/𝒙 =𝝅/𝟐 " + log 2"