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Example 15 - Show (x - y) dy/dx = x + 2y is homogeneous, solve

Example 15 - Chapter 9 Class 12 Differential Equations - Part 2
Example 15 - Chapter 9 Class 12 Differential Equations - Part 3 Example 15 - Chapter 9 Class 12 Differential Equations - Part 4 Example 15 - Chapter 9 Class 12 Differential Equations - Part 5 Example 15 - Chapter 9 Class 12 Differential Equations - Part 6 Example 15 - Chapter 9 Class 12 Differential Equations - Part 7 Example 15 - Chapter 9 Class 12 Differential Equations - Part 8 Example 15 - Chapter 9 Class 12 Differential Equations - Part 9 Example 15 - Chapter 9 Class 12 Differential Equations - Part 10 Example 15 - Chapter 9 Class 12 Differential Equations - Part 11 Example 15 - Chapter 9 Class 12 Differential Equations - Part 12


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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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.