Check sibling questions

Example 17 - Show 2y e x/y dx + (y - 2x ex/y) dy = 0, particular

Example 17 - Chapter 9 Class 12 Differential Equations - Part 2
Example 17 - Chapter 9 Class 12 Differential Equations - Part 3 Example 17 - Chapter 9 Class 12 Differential Equations - Part 4 Example 17 - Chapter 9 Class 12 Differential Equations - Part 5 Example 17 - Chapter 9 Class 12 Differential Equations - Part 6


Transcript

Example 17 Show that the differential equation 2𝑦𝑒^(π‘₯/𝑦) 𝑑π‘₯+(π‘¦βˆ’2π‘₯𝑒^(π‘₯/𝑦) )𝑑𝑦=0 is homogeneous and find its particular solution , given that, π‘₯=0 when 𝑦=1 2𝑦𝑒^(π‘₯/𝑦) 𝑑π‘₯+(π‘¦βˆ’2π‘₯𝑒^(π‘₯/𝑦) )𝑑𝑦 = 0 Step 1: Find 𝑑π‘₯/𝑑𝑦 2𝑦𝑒^(π‘₯/𝑦) 𝑑π‘₯+(π‘¦βˆ’2π‘₯𝑒^(π‘₯/𝑦) )𝑑𝑦=0 2𝑦𝑒^(π‘₯/𝑦) 𝑑π‘₯=βˆ’(π‘¦βˆ’2π‘₯𝑒^(π‘₯/𝑦) )𝑑𝑦 2𝑦𝑒^(π‘₯/𝑦) 𝑑π‘₯=(2π‘₯𝑒^(π‘₯/𝑦)βˆ’π‘¦)𝑑𝑦 Since the equation is in the form π‘₯/𝑦 , we will take 𝑑π‘₯/𝑑𝑦 Instead of 𝑑𝑦/𝑑π‘₯ 𝑑π‘₯/𝑑𝑦=((2π‘₯𝑒^(π‘₯/𝑦) βˆ’ 𝑦))/(2𝑦𝑒^(π‘₯/𝑦) ) Step 2: Put F(π‘₯ , 𝑦)=𝑑π‘₯/𝑑𝑦 and find F(πœ†π‘₯ ,πœ†π‘¦) F(π‘₯ , 𝑦)= (2π‘₯𝑒^(π‘₯/𝑦) βˆ’ 𝑦)/(2𝑦𝑒^(π‘₯/𝑦) ) Finding F(𝝀𝒙 ,π€π’š) F(πœ†π‘₯ ,πœ†π‘¦)=(2 (πœ†π‘₯) γ€– 𝑒〗^(πœ†π‘₯/πœ†π‘¦ βˆ’πœ†π‘¦))/(2πœ†π‘¦ γ€– 𝑒〗^(πœ†π‘₯/πœ†π‘¦ ) )=πœ†(2π‘₯𝑒^(π‘₯/𝑦) βˆ’ 𝑦)/(πœ† . 2𝑦 𝑒^(π‘₯/𝑦) ) =(2π‘₯𝑒^(π‘₯/𝑦) βˆ’ 𝑦)/(2𝑦 𝑒^(π‘₯/𝑦) ) = F (π‘₯ , 𝑦) So, F(πœ†π‘₯ ,πœ†π‘¦)= F(π‘₯ , 𝑦) = πœ†Β° F(π‘₯ , 𝑦) Thus , F(π‘₯ ,𝑦) is a homogeneous function of degree zero Therefore given differential equation is homogeneous differential equation Step 3: Solving 𝑑π‘₯/𝑑𝑦 by Putting π‘₯=𝑣𝑦 𝑑π‘₯/𝑑𝑦=(2π‘₯ 𝑒^(π‘₯/𝑦) βˆ’ 𝑦)/(2𝑦 𝑒^(π‘₯/𝑦) ) Put π‘₯=𝑣𝑦 Diff. w.r.t. 𝑦 𝑑π‘₯/𝑑𝑦=𝑑/𝑑𝑦 (𝑣𝑦) 𝑑π‘₯/𝑑𝑦=𝑦 . 𝑑𝑣/𝑑𝑦+𝑣 𝑑𝑦/𝑑𝑦 𝑑π‘₯/𝑑𝑦=𝑦 . 𝑑𝑣/𝑑𝑦+𝑣 Putting values of 𝑑π‘₯/𝑑𝑦 and x in (1) 𝑑π‘₯/𝑑𝑦=(2π‘₯𝑒^(π‘₯/𝑦) βˆ’ 𝑦)/(2𝑦 𝑒^(π‘₯" " /𝑦) ) 𝑣+𝑦 𝑑𝑣/𝑑𝑦=(2𝑣 𝑒^𝑣 βˆ’ 1)/(2γ€– 𝑒〗^𝑣 ) 𝑦 𝑑𝑣/𝑑𝑦=(2𝑣 𝑒^𝑣 βˆ’ 1)/(2γ€– 𝑒〗^𝑣 )βˆ’π‘£ (𝑦 𝑑𝑣)/𝑑𝑦=(2𝑣𝑒^𝑣 βˆ’ 1 βˆ’ 2𝑣𝑒^𝑣)/(2γ€– 𝑒〗^𝑣 ) 𝑦 𝑑𝑣/𝑑𝑦=(βˆ’1)/(2γ€– 𝑒〗^𝑣 ) 2γ€– 𝑒〗^𝑣 𝑑𝑣=(βˆ’π‘‘π‘¦)/( 𝑦) Integrating Both Sides ∫1β–’γ€–2γ€– 𝑒〗^𝑣 𝑑𝑣〗=∫1β–’(βˆ’π‘‘π‘¦)/( 𝑦) 2γ€– 𝑒〗^𝑣=βˆ’log⁑|𝑦|+𝑐 Putting back 𝑣=π‘₯/𝑦 2𝑒^(π‘₯/𝑦)=βˆ’π‘™π‘œπ‘”|𝑦|+𝑐 2𝑒^(π‘₯/𝑦)+π‘™π‘œπ‘”|𝑦|=𝑐 …(2) Given that at π‘₯=0 , 𝑦=1 Putting π‘₯=0 and 𝑦=1 in (2) 2𝑒^(0/1)βˆ’π‘™π‘œπ‘”|1|=𝑐 2 Γ—1+0=𝑐 𝑐=2 Put Value of 𝑐 in (2) i.e., 2𝑒^(π‘₯/𝑦)+π‘™π‘œπ‘”|𝑦|=𝐢 πŸπ’†^(𝒙/π’š)+π’π’π’ˆ|π’š|=𝟐" " is the particular solution of given differential equation (As π‘™π‘œπ‘”|1| = 0)

Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.