CBSE Class 12 Sample Paper for 2018 Boards

Class 12
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### Show that (x β 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 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 ππ¦/ππ₯ + Py = Q or ππ₯/ππ¦ + P1 x = Q1, yey dx = (y3 + 2xey) dy (π¦π^π¦)/((π¦3 + 2π₯π^π¦)) = ππ¦/ππ₯ ππ¦/ππ₯ = (π¦π^π¦)/((π¦3 + 2π₯π^π¦)) This is not of the form ππ¦/ππ₯ + Py = Q β΄ We find ππ₯/ππ¦ ππ₯/ππ¦ = ((π¦^3 + 2π₯π^π¦))/(π¦π^π¦ ) ππ₯/ππ¦ = π¦^3/(π¦π^π¦ )+(2π₯π^π¦)/(π¦π^π¦ ) ππ₯/ππ¦ = π¦^2/π^π¦ +2π₯/π¦ ππ₯/ππ¦β2π₯/π¦ = π¦^2/π^π¦ ππ₯/ππ¦+((β2)/π¦)π₯ = π¦^2/π^π¦ Comparing with ππ₯/ππ¦ + P1 x = Q1 Where P1 = (β2)/π¦ & Q1 = π¦^2/π^π¦ Find Integrating factor, IF = π^β«1βγπ1 ππ¦γ = π^β«1β(β2ππ¦)/π¦ = π^(β2β«1βππ¦/π¦) = π^(β2 logβ‘π¦ ) Using a log b = log ba = π^(γlogβ‘π¦γ^(β2) ) = π¦^(β2) = 1/π¦^2 Solution is π₯ (IF) = β«1βγ(π1ΓπΌπΉ)ππ¦+πγ π₯ Γ 1/π¦^2 = β«1βγπ¦^2/π^π¦ Γ1/π¦^2 ππ¦+πγ π₯/π¦^2 = β«1βγ1/π^π¦ ππ¦+πγ π₯/π¦^2 = β«1βγπ^(βπ¦) ππ¦+πγ π₯/π¦^2 = β1 Γπ^(βπ¦)+π π₯/π¦^2 = γβπγ^(βπ¦)+π π₯ = γβπ¦^2 πγ^(βπ¦)+ππ¦^2 We need to find particular solution, Putting x = 0, y = 1 0 = γβ1^2 πγ^(β1)+πγ(1)γ^2 0 = γβπγ^(β1)+π π^(β1) =π c=π^(β1) c=1/π Putting value of c in (2) π₯ = γβπ¦^2 πγ^(βπ¦)+ππ¦^2 π₯ = γβπ¦^2 πγ^(βπ¦)+(1/π) π¦^2 π = γβπ^π πγ^(βπ)+π^π/π Question 19 Show that (x β 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 ππ¦/ππ₯ Step 2: Put F(π₯ , π¦)=ππ¦/ππ₯ & Find F(ππ₯ ,ππ¦) Step 3: Then solve using by putting π¦=π£π₯ Finding ππ¦/ππ₯ (x β y)dy = (x + 2y)dx ππ¦/ππ₯=((π₯ + 2π¦)/(π₯ β π¦)) Now, Putting F(π₯ , π¦)=ππ¦/ππ₯ & Find F(ππ₯ ,ππ¦) Let F(π₯ , π¦)=((π₯ + 2π¦)/(π₯ β π¦)) Finding F(ππ ,ππ) F(ππ₯ ,ππ¦)=(ππ₯ + 2(ππ¦))/(ππ₯ βππ¦) =π(π₯ + 2π¦)/(π (π₯ β π¦) ) =((π₯ + 2π¦))/(π₯ β π¦) = F(π₯ , π¦) Thus , F(ππ₯ ,ππ¦)="F" (π₯ , π¦)" " =πΒ°" F" (π₯ , π¦)" " Thus , "F" (π₯ , π¦)" is Homogeneous function of degree zero" Therefore, the given Differential Equation is Homogeneous differential Equation Step 3: Solving ππ¦/ππ₯ by Putting π¦=π£π₯ ππ¦/ππ₯=((π₯ + 2π¦)/(π₯ β π¦)) Let π¦=π£π₯ So , ππ¦/ππ₯=π(π£π₯)/ππ₯ ππ¦/ππ₯=ππ£/ππ₯ . π₯+π£ ππ₯/ππ₯ ππ¦/ππ₯=ππ£/ππ₯ π₯+π£ Putting ππ¦/ππ₯ πnd π¦/π₯ ππ (π) π.π. ππ¦/ππ₯=(π₯ + 2π¦)/(π₯ β π¦) ππ£/ππ₯ π₯+π£= (π₯ + 2 π£π₯)/(π₯ β π£π₯) ππ£/ππ₯ π₯+π£= π₯(1 + 2π£)/π₯(1 β π£) ππ£/ππ₯ π₯+π£= (1 + 2π£)/(1β π£) ππ£/ππ₯ π₯= (1 + 2π£)/(1β π£)βπ£ ππ£/ππ₯ . π₯= (1 + 2π£ β π£ (1β π£))/(1 βπ£) ππ£/ππ₯ . π₯= (1 + 2π£ β π£ +γ π£γ^2)/(1 β π£) ππ£/ππ₯ . π₯= (γ π£γ^2 + π£ + 1)/(1 β π£) ππ£/ππ₯ π₯=β((γ π£γ^2 + π£ + 1)/(π£ β 1)) ππ£((π£β1)/(π£^(2 )+ π£ + 1))=(β ππ₯)/π₯ Integrating Both Sides β«1βγ(π£ β1)/(π£^2 + π£ + 1) ππ£=β«1β(βππ₯)/π₯γ β«1βγ(π£ β1)/(π£^2 + π£ + 1) ππ£=ββ«1βππ₯/π₯γ β«1βγ((π£ β1) ππ£)/(π£^2 + π£ + 1) ππ£γ=βlogβ‘γ|π₯|γ + π We can write π£^2+π£+1 = π£^2 + 1/2 . 2v + (1/2)^2+1β(1/2)^2 =(π£+1/2)^2 + 1 β 1/4 =(π£+1/2)^2+3/4 Putting π£^2+π£+1=(π£+1/2)^2+3/4 & π£β1=π£+π/πβπ/πβ1 =(π£+1/2)β3/2 β«1β((π£ + 1/2) β 3/2)/((π£ + 1/2)^2+ 3/4) ππ£=βlogβ‘γπ₯+πγ β«1β(π£ + 1/2)/((π£ + 1/2)^2+ 3/4) ππ£β3/2 β«1β1/((π£ + 1/2)^2+ 3/4) ππ£=βlogβ‘γπ₯+πγ Thus, I = I1 β (3 )/2 I2 Solving πΌ1 πΌ1=β«1β((π£ + 1/2))/((π£ + 1/2)^2+ 3/4) ππ£ Put (π£+ 1/2)^2+ 3/4 =π‘ Diff. w.r.t. π£ π((π£ + 1/2)^2+ 3/4)/ππ£=ππ‘/ππ£ 2(π£+1/2)=ππ‘/ππ£ ππ£=ππ‘/2(π£ + 1/2) Putting value of v & dv in I1 πΌ1=β«1β((π£ + 1/2))/π‘ Γππ‘/2(π£ + 1/2) =1/2 β«1βππ‘/π‘ =1/2 logβ‘ |π‘| Putting π‘=(π£+ 1/2)^2+3/4 =1/2 πππ|(π£+ 1/2)^2+3/4| =1/2 πππ|π£^2+2π£ Γ 1/2 + 1/4 + 3/4| I1 =1/2 logβ‘γ |π£^2+π£+1|γ Solving π°π πΌ2=β«1βππ£/((π£ + 1/2)^2+3/4) =β«1βππ£/((π£ + 1/2)^2+(β3/2)^2 ) Put π‘=π£+1/2 Diff. w.r.t. π£ ππ‘/ππ£=1 β ππ‘=ππ£ = β«1βππ‘/(π‘^2 + γ (β3/2)γ^2 ) =1/(β3/2) tan^(β1)β‘γπ‘/(β3/2)γ =2/β3 tan^(β1)β‘γ2(π£ + 1/2)/β3γ =2/β3 tan^(β1)β‘((2π£ + 1)/β3) Hence I = πΌ1β3/2 πΌ2 I =1/2 logβ‘ |π£^2+π£+1|β3/2 Γ2/β3 tan^(β1)β‘((2π£ +1)/β3) I =1/2 logβ‘ |π£^2+π£+1|ββ3 tan^(β1)β‘((2π£ +1)/β3) Replacing v by (π¦ )/π₯ I =1/2 πππ|(π¦/π₯)^2+π¦/π₯+1|ββ3 tan^(β1)β‘((2π¦/π₯ + 1)/β3) I =1/2 πππ|π¦^2/π₯^2 +π¦/π₯+1|ββ3 tan^(β1)β‘((2π¦ + π₯)/(β3 π₯)) Putting Value of I in (2) 1/2 πππ|π¦^2/π₯^2 +π¦/π₯+1|ββ3 tan^(β1)β‘((2π¦ + π₯)/(β3 π₯))=βπππ|π₯|+π 1/2 πππ|π¦^2/π₯^2 +π¦/π₯+1|+πππ|π₯|=β3 tan^(β1)β‘((2π¦ + π₯)/(β3 π₯))+π Multiplying Both Sides By 2 πππ|π¦^2/π₯^2 +π¦/π₯+1|+2 πππ|π₯|=2 β3 tan^(β1)β‘((2π¦ + π₯)/(β3 π₯))+2π πππ|π¦^2/π₯^2 +π¦/π₯+1|+πππ|π₯|^2=2β3 tan^(β1)β‘((2π¦ + π₯)/(β3 π₯))+2π Put 2π=π πππ[|π¦^2/π₯^2 +π¦/π₯+1| Γ |π₯^2 |]=2β3 tan^(β1)β‘((2π¦ + π₯)/(β3 π₯))+π πππ|γπ₯^2 π¦γ^2/π₯^2 +(π₯^2 π¦)/π₯+π₯^2 |=2β3 tan^(β1)β‘((π₯ + 2π¦)/(β3 π₯))+π πππ|π^π+ππ+π^π |=πβπ γπππγ^(βπ)β‘((π + ππ)/(βπ π))+π Is the General Solution of the Differential Equation given

#### 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.