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Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


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Example 10 Show that the differential equation (𝑥−𝑦) 𝑑𝑦/𝑑𝑥=𝑥+2𝑦 is homogeneous and solve it.Step 1: Find 𝑑𝑦/𝑑𝑥 (𝑥−𝑦) 𝑑𝑦/𝑑𝑥=𝑥+2𝑦 𝒅𝒚/𝒅𝒙=((𝒙 + 𝟐𝒚)/(𝒙 − 𝒚)) Step 2: Put F(𝑥 , 𝑦)=𝑑𝑦/𝑑𝑥 & Find F(𝜆𝑥 ,𝜆𝑦) 𝑑𝑦/𝑑𝑥=((𝑥 + 2𝑦)/(𝑥 − 𝑦)) Put F(𝑥 , 𝑦)=((𝑥 + 2𝑦)/(𝑥 − 𝑦)) Finding F(𝛌𝐱 ,𝛌𝐲) F(𝜆𝑥 ,𝜆𝑦)=(𝜆𝑥 + 2(𝜆𝑦))/(𝜆𝑥 −𝜆𝑦) =𝜆(𝑥 + 2𝑦)/(𝜆 (𝑥 − 𝑦) ) =((𝑥 + 2𝑦))/(𝑥 − 𝑦) = F(𝑥 , 𝑦) Thus , F(𝜆𝑥 ,𝜆𝑦)="F" (𝑥 , 𝑦)" " =𝝀°" F" (𝒙 , 𝒚)" " Thus , "F" (𝑥 , 𝑦)" is Homogeneous function of degree zero" Therefore, the given Differential Equation is Homogeneous differential Equation Step 3: Solving 𝑑𝑦/𝑑𝑥 by Putting 𝑦=𝑣𝑥 𝑑𝑦/𝑑𝑥=((𝑥 + 2𝑦)/(𝑥 − 𝑦)) Let 𝒚=𝒗𝒙 So , 𝑑𝑦/𝑑𝑥=𝑑(𝑣𝑥)/𝑑𝑥 𝑑𝑦/𝑑𝑥=𝑑𝑣/𝑑𝑥 . 𝑥+𝑣 𝑑𝑥/𝑑𝑥 𝒅𝒚/𝒅𝒙=𝒅𝒗/𝒅𝒙 𝒙+𝒗 Putting 𝑑𝑦/𝑑𝑥 𝑎nd 𝑦/𝑥 𝑖𝑛 (1) 𝑑𝑦/𝑑𝑥=(𝑥 + 2𝑦)/(𝑥 − 𝑦) 𝒅𝒗/𝒅𝒙 𝒙+𝒗= (𝒙 + 𝟐𝒗𝒙)/(𝒙 − 𝒗𝒙) 𝑑𝑣/𝑑𝑥 𝑥+𝑣= 𝑥(1 + 2𝑣)/𝑥(1 − 𝑣) 𝑑𝑣/𝑑𝑥 𝑥+𝑣= (1 + 2𝑣)/(1 − 𝑣) 𝑑𝑣/𝑑𝑥 𝑥= (1 + 2𝑣)/(1 − 𝑣)−𝑣 𝑑𝑣/𝑑𝑥 . 𝑥= (1 + 2𝑣 − 𝑣 (1 − 𝑣))/(1 − 𝑣) 𝑑𝑣/𝑑𝑥 . 𝑥= (1 + 2𝑣 − 𝑣 +〖 𝑣〗^2)/(1 − 𝑣) 𝑑𝑣/𝑑𝑥 . 𝑥= (〖 𝑣〗^2 + 𝑣 + 1)/(1 − 𝑣) 𝑑𝑣/𝑑𝑥 𝑥=−((〖 𝑣〗^2 + 𝑣 + 1)/(𝑣 − 1)) 𝒅𝒗((𝒗 − 𝟏)/(𝒗^(𝟐 )+ 𝒗 + 𝟏))=(−𝒅𝒙)/𝒙 Integrating Both Sides ∫1▒〖(𝑣 − 1)/(𝑣^2 + 𝑣 + 1) 𝑑𝑣=∫1▒(−𝑑𝑥)/𝑥〗 ∫1▒〖(𝑣 − 1)/(𝑣^2 + 𝑣 + 1) 𝑑𝑣=−∫1▒𝑑𝑥/𝑥〗 ∫1▒〖((𝒗 − 𝟏) 𝒅𝒗)/(𝒗^𝟐 + 𝒗 + 𝟏) 𝒅𝒗〗=−𝐥𝐨𝐠⁡〖|𝒙|〗 + 𝒄 We can write 𝑣^2+𝑣+1 = 𝑣^2 + 1/2 . 2v + (1/2)^2+1−(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 𝒗−𝟏=𝑣+𝟏/𝟐−𝟏/𝟐−1 =(𝑣+1/2)−3/2 ∫1▒((𝑣 + 1/2) − 3/2)/((𝑣 + 1/2)^2+ 3/4) 𝑑𝑣=−log⁡〖|𝑥|+𝑐〗 ∫1▒(𝑣 + 1/2)/((𝑣 + 1/2)^2+ 3/4) 𝑑𝑣−3/2 ∫1▒1/((𝑣 + 1/2)^2+ 3/4) 𝑑𝑣=−log⁡〖|𝑥|+𝑐〗 So, our equation becomes I1 – 𝟑/𝟐I2 = −log⁡〖|𝑥|〗+𝑐 Solving 𝑰𝟏 𝐼1=∫1▒((𝑣 + 1/2))/((𝑣 + 1/2)^2+ 3/4) 𝑑𝑣 Put (𝒗+ 𝟏/𝟐)^𝟐+ 𝟑/𝟒 =𝒕 Diff. w.r.t. 𝑣 𝑑((𝑣 + 1/2)^2+ 3/4)/𝑑𝑣=𝑑𝑡/𝑑𝑣 2(𝑣+1/2)=𝑑𝑡/𝑑𝑣 𝑑𝑣=𝑑𝑡/2(𝑣 + 1/2) Putting value of v & dv in I1 𝐼1=∫1▒((𝑣 + 1/2))/𝑡 ×𝑑𝑡/2(𝑣 + 1/2) =1/2 ∫1▒𝑑𝑡/𝑡 =1/2 log⁡ |𝑡| Putting back 𝑡=(𝑣+ 1/2)^2+3/4 =1/2 𝑙𝑜𝑔|(𝑣+ 1/2)^2+3/4| =1/2 𝑙𝑜𝑔|𝑣^2+2𝑣 × 1/2 + 1/4 + 3/4| =𝟏/𝟐 𝒍𝒐𝒈⁡〖 |𝒗^𝟐+𝒗+𝟏|〗 Solving 𝑰𝟐 𝑰𝟐=∫1▒𝑑𝑣/((𝑣 + 1/2)^2+3/4) =∫1▒𝒅𝒗/((𝒗 + 𝟏/𝟐)^𝟐+(√𝟑/𝟐)^𝟐 ) Using ∫1▒〖𝑑𝑥/(𝑥^2 + 𝑎^2 )=(1 )/𝑎 〖𝑡𝑎𝑛〗^(−1)⁡〖𝑥/𝑎〗 〗 where x = v + 1/2 and a = √3/2 =1/(√3/2) tan^(−1)⁡〖((𝑣 + 1/2))/(√3/2)〗 =2/√3 tan^(−1)⁡〖2(𝑣 + 1/2)/√3〗 =𝟐/√𝟑 〖𝒕𝒂𝒏〗^(−𝟏)⁡((𝟐𝒗 + 𝟏)/√𝟑) From (2) I1 – 𝟑/𝟐I2 = −log⁡〖|𝑥|+𝑐〗 1/2 log⁡ |𝑣^2+𝑣+1|−3/2 ×2/√3 tan^(−1)⁡((2𝑣 +1)/√3) = −log⁡〖|𝑥|+𝑐〗 1/2 log⁡ |𝑣^2+𝑣+1|−√3 tan^(−1)⁡((2𝑣 +1)/√3) = −log⁡〖|𝑥|+𝑐〗 Replacing v by (𝑦 )/𝑥 1/2 𝑙𝑜𝑔|(𝑦/𝑥)^2+𝑦/𝑥+1|−√3 tan^(−1)⁡((2𝑦/𝑥 + 1)/√3) = −log⁡〖|𝑥|+𝑐〗 1/2 𝑙𝑜𝑔|𝑦^2/𝑥^2 +𝑦/𝑥+1|−√3 tan^(−1)⁡((2𝑦 + 𝑥)/(√3 𝑥))=−𝑙𝑜𝑔|𝑥|+𝑐 1/2 𝑙𝑜𝑔|𝑦^2/𝑥^2 +𝑦/𝑥+1|+𝑙𝑜𝑔|𝑥|=√3 tan^(−1)⁡((2𝑦 + 𝑥)/(√3 𝑥))+𝑐 Multiplying Both Sides By 2 𝒍𝒐𝒈|𝒚^𝟐/𝒙^𝟐 +𝒚/𝒙+𝟏|+𝟐 𝒍𝒐𝒈|𝒙|=𝟐 √𝟑 〖𝒕𝒂𝒏〗^(−𝟏)⁡((𝟐𝒚 + 𝒙)/(√𝟑 𝒙))+𝟐𝒄 𝑙𝑜𝑔|𝑦^2/𝑥^2 +𝑦/𝑥+1|+𝑙𝑜𝑔|𝑥|^2=2√3 tan^(−1)⁡((2𝑦 + 𝑥)/(√3 𝑥))+2𝑐 Put 2𝑐=𝑐 𝒍𝒐𝒈[|𝒚^𝟐/𝒙^𝟐 +𝒚/𝒙+𝟏| × |𝒙^𝟐 |]=𝟐√𝟑 〖𝒕𝒂𝒏〗^(−𝟏)⁡((𝟐𝒚 + 𝒙)/(√𝟑 𝒙))+𝒄 𝑙𝑜𝑔|〖𝑥^2 𝑦〗^2/𝑥^2 +(𝑥^2 𝑦)/𝑥+𝑥^2 |=2√3 tan^(−1)⁡((𝑥 + 2𝑦)/(√3 𝑥))+𝑐 𝒍𝒐𝒈|𝒙^𝟐+𝒙𝒚+𝒚^𝟐 |=𝟐√𝟑 〖𝒕𝒂𝒏〗^(−𝟏)⁡((𝒙 + 𝟐𝒚)/(√𝟑 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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.