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Example 16 - Show x - cos (y/x) = y cos(y/x) + x is homogeneous - Solving homogeneous differential equation

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  1. Chapter 9 Class 12 Differential Equations
  2. Serial order wise
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Example 16 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‘₯ 𝑑𝑦/𝑑π‘₯=(𝑦 cos⁑ (𝑦/π‘₯) + π‘₯)/(π‘₯ cos⁑(𝑦/π‘₯) ) 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(πœ†π‘₯ ,πœ†π‘¦) 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 (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, 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 𝑦=𝑣π‘₯ 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/𝑑π‘₯ π‘₯+𝑣=((𝑣π‘₯) π‘π‘œπ‘ (𝑣π‘₯/π‘₯) + π‘₯)/(π‘₯ cos⁑(𝑣π‘₯/π‘₯) ) 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=𝑑π‘₯/π‘₯ Integrating Both Sides ∫1β–’cos⁑〖𝑣 𝑑𝑣=∫1▒𝑑π‘₯/π‘₯γ€— sin⁑〖𝑣=log⁑|π‘₯|+𝑐1γ€— Putting 𝑣=𝑦/π‘₯ & t 𝑐1=log⁑𝑐 𝑠𝑖𝑛 𝑦/π‘₯=log⁑〖|π‘₯|+log⁑|𝑐| γ€— π’”π’Šπ’ π’š/𝒙=π’π’π’ˆβ‘|𝒄𝒙|

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