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Example 11 Show that the differential equation π‘₯βˆ’π‘π‘œπ‘ (𝑦/π‘₯)=𝑦 π‘π‘œπ‘ (𝑦/π‘₯)+π‘₯ is homogeneous and solve it.Step 1: Find 𝑑𝑦/𝑑π‘₯ π‘₯ π‘π‘œπ‘ (𝑦/π‘₯) 𝑑𝑦/𝑑π‘₯=𝑦 cos⁑(𝑦/π‘₯)+π‘₯ π’…π’š/𝒅𝒙=(π’š 𝐜𝐨𝐬⁑ (π’š/𝒙) + 𝒙)/(𝒙 𝐜𝐨𝐬⁑(π’š/𝒙) ) Step 2: Put F(π‘₯ ,𝑦)=𝑑𝑦/𝑑π‘₯ & find F(πœ†π‘₯ ,πœ†π‘¦) F(π‘₯ ,𝑦)=(𝑦 cos⁑ (𝑦/π‘₯) + π‘₯)/(π‘₯ cos⁑(𝑦/π‘₯) ) Finding F(𝝀𝒙 ,π€π’š) F(πœ†π‘₯ ,πœ†π‘¦)=((πœ†π‘¦)π‘π‘œπ‘ (πœ†π‘¦/πœ†π‘₯) + πœ†π‘₯)/((πœ†π‘¦) . cos⁑(πœ†π‘¦/πœ†π‘₯) ) =(πœ†π‘¦ π‘π‘œπ‘ (𝑦/π‘₯) + πœ†π‘₯)/(πœ†π‘¦ cos⁑(𝑦/π‘₯) ) =πœ†(𝑦 π‘π‘œπ‘ (𝑦/π‘₯) + π‘₯)/(πœ† π‘₯ cos⁑(𝑦/π‘₯) ) =(𝑦 π‘π‘œπ‘ (𝑦/π‘₯) + π‘₯)/( π‘₯ cos⁑(𝑦/π‘₯) ) = F (𝒙 , π’š) 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, the given differential equation is homogeneous differential equation Step 3: Solving 𝑑𝑦/𝑑π‘₯ by Putting 𝑦=𝑣π‘₯ 𝑑𝑦/𝑑π‘₯=(𝑦 π‘π‘œπ‘ (𝑦/π‘₯) + π‘₯)/(π‘₯ cos⁑(𝑦/π‘₯) ) Put π’š=𝒗𝒙 So, 𝑑𝑦/𝑑π‘₯=𝑑(𝑣π‘₯) =𝑑𝑣/𝑑π‘₯ . π‘₯+𝑣 𝑑π‘₯/𝑑π‘₯ =𝑑𝑣/𝑑π‘₯ π‘₯+𝑣 Putting values of 𝑑𝑦/𝑑π‘₯ and y = vx in (1) i.e. 𝑑𝑦/𝑑π‘₯ = (𝑦 π‘π‘œπ‘ (𝑦/π‘₯)+π‘₯)/(π‘₯ cos⁑(𝑦/π‘₯) ) 𝒅𝒗/𝒅𝒙 𝒙+𝒗=((𝒗𝒙) 𝒄𝒐𝒔(𝒗𝒙/𝒙) + 𝒙)/(𝒙 𝐜𝐨𝐬⁑(𝒗𝒙/𝒙) ) 𝑑𝑣/𝑑π‘₯ π‘₯+𝑣=(𝑣π‘₯ π‘π‘œπ‘ (𝑣) + π‘₯)/(π‘₯ cos⁑𝑣 ) 𝑑𝑣/𝑑π‘₯ π‘₯+𝑣=π‘₯(𝑣 cos⁑〖𝑣 +1γ€— )/(π‘₯ cos⁑𝑣 ) 𝑑𝑣/𝑑π‘₯ π‘₯+𝑣=(𝑣 cos⁑〖𝑣 +1γ€—)/cos⁑𝑣 𝑑𝑣/𝑑π‘₯ π‘₯=(𝑣 cos⁑〖𝑣 + 1γ€—)/cos⁑𝑣 βˆ’π‘£ 𝑑𝑣/𝑑π‘₯ π‘₯=(𝑣 cos⁑〖𝑣 + 1γ€— βˆ’π‘£ cos⁑𝑣)/cos⁑𝑣 𝑑𝑣/𝑑π‘₯ π‘₯= 1/cos⁑𝑣 𝒄𝒐𝒔⁑𝒗 𝒅𝒗=𝒅𝒙/𝒙 Integrating Both Sides ∫1β–’cos⁑〖𝑣 𝑑𝑣=∫1▒𝑑π‘₯/π‘₯γ€— sin⁑〖𝑣=log⁑|π‘₯|+𝑐1γ€— Putting 𝒗=π’š/𝒙 & t π’„πŸ=π₯𝐨𝐠⁑𝒄 𝑠𝑖𝑛 𝑦/π‘₯=log⁑〖|π‘₯|+log⁑|𝑐| γ€— π’”π’Šπ’ π’š/𝒙=π’π’π’ˆβ‘|𝒄𝒙|

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo