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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  1. Class 12
  2. Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

Transcript

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 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.