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Example 10 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 π’…π’š/𝒅𝒙=((𝒙 + πŸπ’š)/(𝒙 βˆ’ π’š)) 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)" " =𝝀°" F" (𝒙 , π’š)" " 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 π’š=𝒗𝒙 So , 𝑑𝑦/𝑑π‘₯=𝑑(𝑣π‘₯)/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯=𝑑𝑣/𝑑π‘₯ . π‘₯+𝑣 𝑑π‘₯/𝑑π‘₯ π’…π’š/𝒅𝒙=𝒅𝒗/𝒅𝒙 𝒙+𝒗 Putting 𝑑𝑦/𝑑π‘₯ π‘Žnd 𝑦/π‘₯ 𝑖𝑛 (1) 𝑑𝑦/𝑑π‘₯=(π‘₯ + 2𝑦)/(π‘₯ βˆ’ 𝑦) 𝒅𝒗/𝒅𝒙 𝒙+𝒗= (𝒙 + πŸπ’—π’™)/(𝒙 βˆ’ 𝒗𝒙) 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’™)/𝒙 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’π₯𝐨𝐠⁑〖|𝒙|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’πŸ=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+ πŸ‘/πŸ’ =𝒕 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| =𝟏/𝟐 π’π’π’ˆβ‘γ€– |𝒗^𝟐+𝒗+𝟏|γ€— Solving π‘°πŸ π‘°πŸ=∫1▒𝑑𝑣/((𝑣 + 1/2)^2+3/4) =∫1▒𝒅𝒗/((𝒗 + 𝟏/𝟐)^𝟐+(βˆšπŸ‘/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/βˆšπŸ‘ 〖𝒕𝒂𝒏〗^(βˆ’πŸ)⁑((πŸπ’— + 𝟏)/βˆšπŸ‘) 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’π’ˆ|π’š^𝟐/𝒙^𝟐 +π’š/𝒙+𝟏|+𝟐 π’π’π’ˆ|𝒙|=𝟐 βˆšπŸ‘ 〖𝒕𝒂𝒏〗^(βˆ’πŸ)⁑((πŸπ’š + 𝒙)/(βˆšπŸ‘ 𝒙))+πŸπ’„ 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’π’ˆ[|π’š^𝟐/𝒙^𝟐 +π’š/𝒙+𝟏| Γ— |𝒙^𝟐 |]=πŸβˆšπŸ‘ 〖𝒕𝒂𝒏〗^(βˆ’πŸ)⁑((πŸπ’š + 𝒙)/(βˆšπŸ‘ 𝒙))+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 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