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)

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.