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Find the particular solution of the differential equation :

ye y dx = (y 3 + 2xe y ) dy,  y(0) = 1

OR

Show that (x − y)dy = (x + 2y)dx is a homogenous differential equation. Also, find the general solution of the given differential equation.

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Question 19 - CBSE Class 12 Sample Paper for 2018 Boards - Part 2
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Question 19 - CBSE Class 12 Sample Paper for 2018 Boards - Part 6 Question 19 - CBSE Class 12 Sample Paper for 2018 Boards - Part 7 Question 19 - CBSE Class 12 Sample Paper for 2018 Boards - Part 8 Question 19 - CBSE Class 12 Sample Paper for 2018 Boards - Part 9 Question 19 - CBSE Class 12 Sample Paper for 2018 Boards - Part 10 Question 19 - CBSE Class 12 Sample Paper for 2018 Boards - Part 11 Question 19 - CBSE Class 12 Sample Paper for 2018 Boards - Part 12 Question 19 - CBSE Class 12 Sample Paper for 2018 Boards - Part 13 Question 19 - CBSE Class 12 Sample Paper for 2018 Boards - Part 14 Question 19 - CBSE Class 12 Sample Paper for 2018 Boards - Part 15 Question 19 - CBSE Class 12 Sample Paper for 2018 Boards - Part 16 Question 19 - CBSE Class 12 Sample Paper for 2018 Boards - Part 17 Question 19 - CBSE Class 12 Sample Paper for 2018 Boards - Part 18 Question 19 - CBSE Class 12 Sample Paper for 2018 Boards - Part 19

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Question 19 Find the particular solution of the differential equation : yey dx = (y3 + 2xey) dy, y(0) = 1 Given equation yey dx = (y3 + 2xey) dy We try to put in form 𝑑𝑦/𝑑π‘₯ + Py = Q or 𝑑π‘₯/𝑑𝑦 + P1 x = Q1, yey dx = (y3 + 2xey) dy (𝑦𝑒^𝑦)/((𝑦3 + 2π‘₯𝑒^𝑦)) = 𝑑𝑦/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯ = (𝑦𝑒^𝑦)/((𝑦3 + 2π‘₯𝑒^𝑦)) This is not of the form 𝑑𝑦/𝑑π‘₯ + Py = Q ∴ We find 𝑑π‘₯/𝑑𝑦 𝑑π‘₯/𝑑𝑦 = ((𝑦^3 + 2π‘₯𝑒^𝑦))/(𝑦𝑒^𝑦 ) 𝑑π‘₯/𝑑𝑦 = 𝑦^3/(𝑦𝑒^𝑦 )+(2π‘₯𝑒^𝑦)/(𝑦𝑒^𝑦 ) 𝑑π‘₯/𝑑𝑦 = 𝑦^2/𝑒^𝑦 +2π‘₯/𝑦 𝑑π‘₯/π‘‘π‘¦βˆ’2π‘₯/𝑦 = 𝑦^2/𝑒^𝑦 𝑑π‘₯/𝑑𝑦+((βˆ’2)/𝑦)π‘₯ = 𝑦^2/𝑒^𝑦 Comparing with 𝑑π‘₯/𝑑𝑦 + P1 x = Q1 Where P1 = (βˆ’2)/𝑦 & Q1 = 𝑦^2/𝑒^𝑦 Find Integrating factor, IF = 𝑒^∫1▒〖𝑃1 𝑑𝑦〗 = 𝑒^∫1β–’(βˆ’2𝑑𝑦)/𝑦 = 𝑒^(βˆ’2∫1▒𝑑𝑦/𝑦) = 𝑒^(βˆ’2 log⁑𝑦 ) Using a log b = log ba = 𝑒^(γ€–log⁑𝑦〗^(βˆ’2) ) = 𝑦^(βˆ’2) = 1/𝑦^2 Solution is π‘₯ (IF) = ∫1β–’γ€–(𝑄1×𝐼𝐹)𝑑𝑦+𝑐〗 π‘₯ Γ— 1/𝑦^2 = ∫1▒〖𝑦^2/𝑒^𝑦 Γ—1/𝑦^2 𝑑𝑦+𝑐〗 π‘₯/𝑦^2 = ∫1β–’γ€–1/𝑒^𝑦 𝑑𝑦+𝑐〗 π‘₯/𝑦^2 = ∫1▒〖𝑒^(βˆ’π‘¦) 𝑑𝑦+𝑐〗 π‘₯/𝑦^2 = βˆ’1 ×𝑒^(βˆ’π‘¦)+𝑐 π‘₯/𝑦^2 = γ€–βˆ’π‘’γ€—^(βˆ’π‘¦)+𝑐 π‘₯ = γ€–βˆ’π‘¦^2 𝑒〗^(βˆ’π‘¦)+𝑐𝑦^2 We need to find particular solution, Putting x = 0, y = 1 0 = γ€–βˆ’1^2 𝑒〗^(βˆ’1)+𝑐〖(1)γ€—^2 0 = γ€–βˆ’π‘’γ€—^(βˆ’1)+𝑐 𝑒^(βˆ’1) =𝑐 c=𝑒^(βˆ’1) c=1/𝑒 Putting value of c in (2) π‘₯ = γ€–βˆ’π‘¦^2 𝑒〗^(βˆ’π‘¦)+𝑐𝑦^2 π‘₯ = γ€–βˆ’π‘¦^2 𝑒〗^(βˆ’π‘¦)+(1/𝑒) 𝑦^2 𝒙 = γ€–βˆ’π’š^𝟐 𝒆〗^(βˆ’π’š)+π’š^𝟐/𝒆 Question 19 Show that (x βˆ’ y)dy = (x + 2y)dx is a homogenous differential equation. Also, find the general solution of the given differential equation. Theory To prove homogenous Step 1: Find 𝑑𝑦/𝑑π‘₯ Step 2: Put F(π‘₯ , 𝑦)=𝑑𝑦/𝑑π‘₯ & Find F(πœ†π‘₯ ,πœ†π‘¦) Step 3: Then solve using by putting 𝑦=𝑣π‘₯ Finding 𝑑𝑦/𝑑π‘₯ (x βˆ’ y)dy = (x + 2y)dx 𝑑𝑦/𝑑π‘₯=((π‘₯ + 2𝑦)/(π‘₯ βˆ’ 𝑦)) Now, Putting F(π‘₯ , 𝑦)=𝑑𝑦/𝑑π‘₯ & Find F(πœ†π‘₯ ,πœ†π‘¦) Let F(π‘₯ , 𝑦)=((π‘₯ + 2𝑦)/(π‘₯ βˆ’ 𝑦)) Finding F(𝝀𝒙 ,π€π’š) 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 (𝑖) 𝑖.𝑒. 𝑑𝑦/𝑑π‘₯=(π‘₯ + 2𝑦)/(π‘₯ βˆ’ 𝑦) 𝑑𝑣/𝑑π‘₯ π‘₯+𝑣= (π‘₯ + 2 𝑣π‘₯)/(π‘₯ βˆ’ 𝑣π‘₯) 𝑑𝑣/𝑑π‘₯ π‘₯+𝑣= π‘₯(1 + 2𝑣)/π‘₯(1 βˆ’ 𝑣) 𝑑𝑣/𝑑π‘₯ π‘₯+𝑣= (1 + 2𝑣)/(1βˆ’ 𝑣) 𝑑𝑣/𝑑π‘₯ π‘₯= (1 + 2𝑣)/(1βˆ’ 𝑣)βˆ’π‘£ 𝑑𝑣/𝑑π‘₯ . π‘₯= (1 + 2𝑣 βˆ’ 𝑣 (1βˆ’ 𝑣))/(1 βˆ’π‘£) 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((π‘£βˆ’1)/(𝑣^(2 )+ 𝑣 + 1))=(βˆ’ 𝑑π‘₯)/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 & 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〗 Thus, I = I1 – (3 )/2 I2 Solving 𝐼1 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 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 ) Put 𝑑=𝑣+1/2 Diff. w.r.t. 𝑣 𝑑𝑑/𝑑𝑣=1 β‡’ 𝑑𝑑=𝑑𝑣 = ∫1▒𝑑𝑑/(𝑑^2 + γ€– (√3/2)γ€—^2 ) =1/(√3/2) tan^(βˆ’1)⁑〖𝑑/(√3/2)γ€— =2/√3 tan^(βˆ’1)⁑〖2(𝑣 + 1/2)/√3γ€— =2/√3 tan^(βˆ’1)⁑((2𝑣 + 1)/√3) Hence I = 𝐼1βˆ’3/2 𝐼2 I =1/2 log⁑ |𝑣^2+𝑣+1|βˆ’3/2 Γ—2/√3 tan^(βˆ’1)⁑((2𝑣 +1)/√3) I =1/2 log⁑ |𝑣^2+𝑣+1|βˆ’βˆš3 tan^(βˆ’1)⁑((2𝑣 +1)/√3) Replacing v by (𝑦 )/π‘₯ I =1/2 π‘™π‘œπ‘”|(𝑦/π‘₯)^2+𝑦/π‘₯+1|βˆ’βˆš3 tan^(βˆ’1)⁑((2𝑦/π‘₯ + 1)/√3) I =1/2 π‘™π‘œπ‘”|𝑦^2/π‘₯^2 +𝑦/π‘₯+1|βˆ’βˆš3 tan^(βˆ’1)⁑((2𝑦 + π‘₯)/(√3 π‘₯)) Putting Value of I in (2) 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 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 Is the General Solution of the Differential Equation given

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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.