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  1. Chapter 9 Class 12 Differential Equations
  2. Serial order wise

Transcript

Example 15 Show that the differential equation (π‘₯βˆ’π‘¦) 𝑑𝑦/𝑑π‘₯=π‘₯+2𝑦 is homogeneous and solve it. Step 1: Find 𝑑𝑦/𝑑π‘₯ (π‘₯βˆ’π‘¦) 𝑑𝑦/𝑑π‘₯=π‘₯+2𝑦 𝑑𝑦/𝑑π‘₯=((π‘₯ + 2𝑦)/(π‘₯ βˆ’ 𝑦)) Step 2: Put F(π‘₯ , 𝑦)=𝑑𝑦/𝑑π‘₯ & Find F(πœ†π‘₯ ,πœ†π‘¦) 𝑑𝑦/𝑑π‘₯=((π‘₯ + 2𝑦)/(π‘₯ βˆ’ 𝑦)) Put F(π‘₯ , 𝑦)=((π‘₯ + 2𝑦)/(π‘₯ βˆ’ 𝑦)) Finding F(π›Œπ± ,π›Œπ²) F(πœ†π‘₯ ,πœ†π‘¦)=(πœ†π‘₯ + 2(πœ†π‘¦))/(πœ†π‘₯ βˆ’πœ†π‘¦) =πœ†(π‘₯ + 2𝑦)/(πœ† (π‘₯ βˆ’ 𝑦) ) =((π‘₯ + 2𝑦))/(π‘₯ βˆ’ 𝑦) = F(π‘₯ , 𝑦) Thus , F(πœ†π‘₯ ,πœ†π‘¦)="F" (π‘₯ , 𝑦)" " =πœ†Β°" F" (π‘₯ , 𝑦)" " Thus , "F" (π‘₯ , 𝑦)" is 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‘₯=((π‘₯ + 2𝑦)/(π‘₯ βˆ’ 𝑦)) Let 𝑦=𝑣π‘₯ So , 𝑑𝑦/𝑑π‘₯=𝑑(𝑣π‘₯)/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯=𝑑𝑣/𝑑π‘₯ . π‘₯+𝑣 𝑑π‘₯/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯=𝑑𝑣/𝑑π‘₯ π‘₯+𝑣 Putting 𝑑𝑦/𝑑π‘₯ π‘Žnd 𝑦/π‘₯ 𝑖𝑛 (1) 𝑑𝑦/𝑑π‘₯=(π‘₯ + 2𝑦)/(π‘₯ βˆ’ 𝑦) 𝑑𝑣/𝑑π‘₯ π‘₯+𝑣= (π‘₯ + 2𝑣π‘₯)/(π‘₯ βˆ’ 𝑣π‘₯) 𝑑𝑣/𝑑π‘₯ π‘₯+𝑣= π‘₯(1 + 2𝑣)/π‘₯(1 βˆ’ 𝑣) 𝑑𝑣/𝑑π‘₯ π‘₯+𝑣= (1 + 2𝑣)/(1 βˆ’ 𝑣) 𝑑𝑣/𝑑π‘₯ π‘₯= (1 + 2𝑣)/(1 βˆ’ 𝑣)βˆ’π‘£ 𝑑𝑣/𝑑π‘₯ . π‘₯= (1 + 2𝑣 βˆ’ 𝑣 (1 βˆ’ 𝑣))/(1 βˆ’ 𝑣) 𝑑𝑣/𝑑π‘₯ . π‘₯= (1 + 2𝑣 βˆ’ 𝑣 +γ€– 𝑣〗^2)/(1 βˆ’ 𝑣) 𝑑𝑣/𝑑π‘₯ . π‘₯= (γ€– 𝑣〗^2 + 𝑣 + 1)/(1 βˆ’ 𝑣) 𝑑𝑣/𝑑π‘₯ π‘₯=βˆ’((γ€– 𝑣〗^2 + 𝑣 + 1)/(𝑣 βˆ’ 1)) 𝑑𝑣((𝑣 βˆ’ 1)/(𝑣^(2 )+ 𝑣 + 1))=(βˆ’π‘‘π‘₯)/π‘₯ Integrating Both Sides ∫1β–’γ€–(𝑣 βˆ’ 1)/(𝑣^2 + 𝑣 + 1) 𝑑𝑣=∫1β–’(βˆ’π‘‘π‘₯)/π‘₯γ€— ∫1β–’γ€–(𝑣 βˆ’ 1)/(𝑣^2 + 𝑣 + 1) 𝑑𝑣=βˆ’βˆ«1▒𝑑π‘₯/π‘₯γ€— ∫1β–’γ€–((𝑣 βˆ’ 1) 𝑑𝑣)/(𝑣^2 + 𝑣 + 1) 𝑑𝑣〗=βˆ’log⁑〖|π‘₯|γ€— + 𝑐 We can write 𝑣^2+𝑣+1 = 𝑣^2 + 1/2 . 2v + (1/2)^2+1βˆ’(1/2)^2 =(𝑣+1/2)^2 + 1 – 1/4 =(𝑣+1/2)^2+3/4 Putting 𝑣^2+𝑣+1=(𝑣+1/2)^2+3/4 and π‘£βˆ’1=𝑣+𝟏/πŸβˆ’πŸ/πŸβˆ’1 =(𝑣+1/2)βˆ’3/2 ∫1β–’((𝑣 + 1/2) βˆ’ 3/2)/((𝑣 + 1/2)^2+ 3/4) 𝑑𝑣=βˆ’log⁑〖|π‘₯|+𝑐〗 ∫1β–’(𝑣 + 1/2)/((𝑣 + 1/2)^2+ 3/4) π‘‘π‘£βˆ’3/2 ∫1β–’1/((𝑣 + 1/2)^2+ 3/4) 𝑑𝑣=βˆ’log⁑〖|π‘₯|+𝑐〗 So, our equation becomes I1 – πŸ‘/𝟐I2 = βˆ’log⁑〖|π‘₯|γ€—+𝑐 Solving π‘°πŸ 𝐼1=∫1β–’((𝑣 + 1/2))/((𝑣 + 1/2)^2+ 3/4) 𝑑𝑣 Put (𝑣+ 1/2)^2+ 3/4 =𝑑 Diff. w.r.t. 𝑣 𝑑((𝑣 + 1/2)^2+ 3/4)/𝑑𝑣=𝑑𝑑/𝑑𝑣 2(𝑣+1/2)=𝑑𝑑/𝑑𝑣 𝑑𝑣=𝑑𝑑/2(𝑣 + 1/2) Putting value of v & dv in I1 𝐼1=∫1β–’((𝑣 + 1/2))/𝑑 ×𝑑𝑑/2(𝑣 + 1/2) =1/2 ∫1▒𝑑𝑑/𝑑 =1/2 log⁑ |𝑑| Putting back 𝑑=(𝑣+ 1/2)^2+3/4 =1/2 π‘™π‘œπ‘”|(𝑣+ 1/2)^2+3/4| =1/2 π‘™π‘œπ‘”|𝑣^2+2𝑣 Γ— 1/2 + 1/4 + 3/4| I1 =1/2 log⁑〖 |𝑣^2+𝑣+1|γ€— Solving π‘°πŸ 𝐼2=∫1▒𝑑𝑣/((𝑣 + 1/2)^2+3/4) =∫1▒𝑑𝑣/((𝑣 + 1/2)^2+(√3/2)^2 ) Using ∫1▒〖𝑑π‘₯/(π‘₯^2 + π‘Ž^2 )=(1 )/π‘Ž γ€–π‘‘π‘Žπ‘›γ€—^(βˆ’1)⁑〖π‘₯/π‘Žγ€— γ€— where x = v + 1/2 and a = √3/2 =1/(√3/2) tan^(βˆ’1)⁑〖((𝑣 + 1/2))/(√3/2)γ€— =2/√3 tan^(βˆ’1)⁑〖2(𝑣 + 1/2)/√3γ€— =2/√3 tan^(βˆ’1)⁑((2𝑣 + 1)/√3) From (2) I1 – πŸ‘/𝟐I2 = βˆ’log⁑〖|π‘₯|+𝑐〗 1/2 log⁑ |𝑣^2+𝑣+1|βˆ’3/2 Γ—2/√3 tan^(βˆ’1)⁑((2𝑣 +1)/√3) = βˆ’log⁑〖|π‘₯|+𝑐〗 1/2 log⁑ |𝑣^2+𝑣+1|βˆ’βˆš3 tan^(βˆ’1)⁑((2𝑣 +1)/√3) = βˆ’log⁑〖|π‘₯|+𝑐〗 Replacing v by (𝑦 )/π‘₯ 1/2 π‘™π‘œπ‘”|(𝑦/π‘₯)^2+𝑦/π‘₯+1|βˆ’βˆš3 tan^(βˆ’1)⁑((2𝑦/π‘₯ + 1)/√3) = βˆ’log⁑〖|π‘₯|+𝑐〗 1/2 π‘™π‘œπ‘”|𝑦^2/π‘₯^2 +𝑦/π‘₯+1|βˆ’βˆš3 tan^(βˆ’1)⁑((2𝑦 + π‘₯)/(√3 π‘₯))=βˆ’π‘™π‘œπ‘”|π‘₯|+𝑐 1/2 π‘™π‘œπ‘”|𝑦^2/π‘₯^2 +𝑦/π‘₯+1|+π‘™π‘œπ‘”|π‘₯|=√3 tan^(βˆ’1)⁑((2𝑦 + π‘₯)/(√3 π‘₯))+𝑐 Multiplying Both Sides By 2 π‘™π‘œπ‘”|𝑦^2/π‘₯^2 +𝑦/π‘₯+1|+2 π‘™π‘œπ‘”|π‘₯|=2 √3 tan^(βˆ’1)⁑((2𝑦 + π‘₯)/(√3 π‘₯))+2𝑐 π‘™π‘œπ‘”|𝑦^2/π‘₯^2 +𝑦/π‘₯+1|+π‘™π‘œπ‘”|π‘₯|^2=2√3 tan^(βˆ’1)⁑((2𝑦 + π‘₯)/(√3 π‘₯))+2𝑐 π‘™π‘œπ‘”|𝑦^2/π‘₯^2 +𝑦/π‘₯+1|+π‘™π‘œπ‘”|π‘₯|^2=2√3 tan^(βˆ’1)⁑((2𝑦 + π‘₯)/(√3 π‘₯))+2𝑐 Put 2𝑐=𝑐 π‘™π‘œπ‘”[|𝑦^2/π‘₯^2 +𝑦/π‘₯+1| Γ— |π‘₯^2 |]=2√3 tan^(βˆ’1)⁑((2𝑦 + π‘₯)/(√3 π‘₯))+𝑐 π‘™π‘œπ‘”|γ€–π‘₯^2 𝑦〗^2/π‘₯^2 +(π‘₯^2 𝑦)/π‘₯+π‘₯^2 |=2√3 tan^(βˆ’1)⁑((π‘₯ + 2𝑦)/(√3 π‘₯))+𝑐 π’π’π’ˆ|𝒙^𝟐+π’™π’š+π’š^𝟐 |=πŸβˆšπŸ‘ 〖𝒕𝒂𝒏〗^(βˆ’πŸ)⁑((𝒙 + πŸπ’š)/(βˆšπŸ‘ 𝒙))+𝒄

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.