Ex 9.5, 10 - Chapter 9 Class 12 Differential Equations (Term 2)

Transcript

Ex 9.5, 10 In each of the Exercise 1 to 10 , show that the given differential equation is homogeneous and solve each of them. (1+𝑒^(𝑥/𝑦) )𝑑𝑥+𝑒^(𝑥/𝑦) (1−𝑥/𝑦)𝑑𝑦=0 Since the equation is in the form 𝑥/𝑦 , we will take 𝑑𝑥/𝑑𝑦 Instead of 𝑑𝑦/𝑑𝑥 Step 1 : Find 𝑑𝑥/𝑑𝑦 (1+𝑒^(𝑥/𝑦) )𝑑𝑥+𝑒^(𝑥/𝑦) (1−𝑥/𝑦)𝑑𝑦 = 0 (1+𝑒^(𝑥/𝑦) ) dx = −𝑒^(𝑥/𝑦) (1−𝑥/𝑦)𝑑𝑦 𝑑𝑥/𝑑𝑦 = (−𝑒^(𝑥/𝑦) (1 − 𝑥/𝑦) )/(1 + 𝑒^(𝑥/𝑦) ) …(1) Step 2: Put 𝑑𝑥/𝑑𝑦 = F(x, y) and find F(𝜆x, 𝜆y) F(x, y) = (−𝑒^(𝑥/𝑦) (1 − 𝑥/𝑦) )/(1 + 𝑒^(𝑥/𝑦) ) F(𝜆x, 𝜆y) = (−𝑒^(𝜆𝑥/𝜆𝑦) (1 − 𝜆𝑥/𝜆𝑦) )/(1 + 𝑒^(𝜆𝑥/𝜆𝑦) ) = (−𝑒^(𝑥/𝑦) (1 − 𝑥/𝑦) )/(1 + 𝑒^(𝑥/𝑦) ) = 𝐹(𝑥, 𝑦) So, F(𝜆𝑥 ,𝜆𝑦)= F(𝑥 , 𝑦) = 𝜆° F(𝑥 , 𝑦) Thus , F(𝑥 ,𝑦) is a homogeneous function of degree zero Therefore given differential equation is homogeneous differential equation Step 3: Solving 𝑑𝑥/𝑑𝑦 by Putting 𝑥=𝑣𝑦 Putting 𝑥=𝑣𝑦 Diff. w.r.t. 𝑦 𝑑𝑥/𝑑𝑦=𝑑/𝑑𝑦 (𝑣𝑦) 𝑑𝑥/𝑑𝑦=𝑦 . 𝑑𝑣/𝑑𝑦+𝑣 𝑑𝑦/𝑑𝑦 𝑑𝑥/𝑑𝑦=𝑦 . 𝑑𝑣/𝑑𝑦+𝑣 Putting values of 𝑑𝑥/𝑑𝑦 and x in (1) 𝑑𝑥/𝑑𝑦=(−𝑒^(𝑥/𝑦) (1 − 𝑥/𝑦) )/(1 + 𝑒^(𝑥/𝑦) ) 𝑣+𝑦 𝑑𝑣/𝑑𝑦=(−𝑒^𝑣 (1 − 𝑣))/(1 + 𝑒^𝑣 ) 𝑦 𝑑𝑣/𝑑𝑦=(−𝑒^𝑣 (1 − 𝑣))/(1 + 𝑒^𝑣 )−𝑣 𝑦 𝑑𝑣/𝑑𝑦=(−𝑒^𝑣+ 𝑣𝑒^𝑣)/(1 + 𝑒^𝑣 )−𝑣 𝑦 𝑑𝑣/𝑑𝑦=(−𝑒^𝑣+ 𝑣𝑒^𝑣 − 𝑣(1 + 𝑒^𝑣 ))/(1 + 𝑒^𝑣 ) 𝑦 𝑑𝑣/𝑑𝑦=(−𝑒^𝑣+ 𝑣𝑒^𝑣 − 𝑣 − 𝑣𝑒^𝑣)/(1 + 𝑒^𝑣 ) 𝑦 𝑑𝑣/𝑑𝑦=(−𝑒^𝑣− 𝑣)/(1 + 𝑒^𝑣 ) 𝑦 𝑑𝑣/𝑑𝑦=(−(𝑒^𝑣+ 𝑣))/(1 + 𝑒^𝑣 ) 𝑦 𝑑𝑣/𝑑𝑦=(−(𝑒^𝑣+ 𝑣))/(1 + 𝑒^𝑣 ) 〖1 + 𝑒〗^𝑣/(𝑣 + 𝑒^𝑣 ) 𝑑𝑣 = (−𝑑𝑦)/𝑦 Integrating both sides ∫1▒〖〖1 + 𝑒〗^𝑣/(𝑣 + 𝑒^𝑣 ) 𝑑𝑣" " 〗 =∫1▒(−𝑑𝑦)/𝑦 ∫1▒〖〖1 + 𝑒〗^𝑣/(𝑣 + 𝑒^𝑣 ) 𝑑𝑣〗=−log⁡〖|𝑦|〗+log⁡𝑐 Putting v + ev = t (1 + ev) dv = dt Thus, our equation becomes ∫1▒𝑑𝑡/𝑡=−log⁡〖|𝑦|〗+log⁡𝑐 log⁡〖|𝑡|〗=−log⁡〖|𝑦|〗+log⁡𝑐 Putting back value of t = v + ev log⁡〖|𝑣+𝑒^𝑣 |〗=−log⁡〖|𝑦|〗+log⁡𝑐 log⁡〖|𝑣+𝑒^𝑣 |〗+log⁡〖|𝑦|〗=log⁡𝑐 log⁡(|𝑣+𝑒^𝑣 |×|𝑦|)=log⁡𝑐 log⁡((𝑣+𝑒^𝑣 )×𝑦)=log⁡𝑐 log⁡(𝑣𝑦+𝑒^𝑣 𝑦)=log⁡𝑐 Putting back value of v = 𝑥/𝑦 log⁡(𝑥/𝑦×𝑦+𝑒^(𝑥/𝑦) 𝑦)=log⁡𝑐 log⁡(𝑥+𝑒^(𝑥/𝑦) 𝑦)=log⁡𝑐 Canceling log 𝒙+𝒚𝒆^(𝒙/𝒚)=𝑪