Examples

Chapter 9 Class 12 Differential Equations
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Example 10 Show that the differential equation (π₯βπ¦) ππ¦/ππ₯=π₯+2π¦ is homogeneous and solve it.Step 1: Find ππ¦/ππ₯ (π₯βπ¦) ππ¦/ππ₯=π₯+2π¦ ππ/ππ=((π + ππ)/(π β π)) Step 2: Put F(π₯ , π¦)=ππ¦/ππ₯ & Find F(ππ₯ ,ππ¦) ππ¦/ππ₯=((π₯ + 2π¦)/(π₯ β π¦)) Put F(π₯ , π¦)=((π₯ + 2π¦)/(π₯ β π¦)) Finding F(ππ± ,ππ²) 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 (1) ππ¦/ππ₯=(π₯ + 2π¦)/(π₯ β π¦) ππ/ππ π+π= (π + π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£)/(1 β π£) ππ£/ππ₯ π₯= (1 + 2π£)/(1 β π£)βπ£ ππ£/ππ₯ . π₯= (1 + 2π£ β π£ (1 β 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 π)/(π^(π )+ π + π))=(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 π) ππ)/(π^π + π + π) ππγ=βπ₯π¨π β‘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^π+π+π=(π£+1/2)^2+3/4 and πβπ=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 So, our equation becomes I1 β π/πI2 = βlogβ‘γ|π₯|γ+π Solving π°π πΌ1=β«1β((π£ + 1/2))/((π£ + 1/2)^2+ 3/4) ππ£ Put (π+ π/π)^π+ π/π =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 back 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| =π/π πππβ‘γ |π^π+π+π|γ Solving π°π π°π=β«1βππ£/((π£ + 1/2)^2+3/4) =β«1βππ/((π + π/π)^π+(βπ/π)^π ) Using β«1βγππ₯/(π₯^2 + π^2 )=(1 )/π γπ‘ππγ^(β1)β‘γπ₯/πγ γ where x = v + 1/2 and a = β3/2 =1/(β3/2) tan^(β1)β‘γ((π£ + 1/2))/(β3/2)γ =2/β3 tan^(β1)β‘γ2(π£ + 1/2)/β3γ =π/βπ γπππγ^(βπ)β‘((ππ + π)/βπ) From (2) I1 β π/πI2 = βlogβ‘γ|π₯|+πγ 1/2 logβ‘ |π£^2+π£+1|β3/2 Γ2/β3 tan^(β1)β‘((2π£ +1)/β3) = βlogβ‘γ|π₯|+πγ 1/2 logβ‘ |π£^2+π£+1|ββ3 tan^(β1)β‘((2π£ +1)/β3) = βlogβ‘γ|π₯|+πγ Replacing v by (π¦ )/π₯ 1/2 πππ|(π¦/π₯)^2+π¦/π₯+1|ββ3 tan^(β1)β‘((2π¦/π₯ + 1)/β3) = βlogβ‘γ|π₯|+πγ 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^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π 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^π/π^π +π/π+π| Γ |π^π |]=πβπ γπππγ^(βπ)β‘((ππ + π)/(βπ 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