Ex 9.4

Chapter 9 Class 12 Differential Equations
Serial order wise

### Transcript

Ex 9.4, 1 In each of the Exercise 1 to 10 , show that the given differential equation is homogeneous and solve (x^2+xy)dy=(x^2+y^2 )πx (π₯^2+π₯π¦)ππ¦ =(π₯^2+π¦^2 )ππ₯ Step 1: Find ππ¦/ππ₯ (x2 + xy)dy = (π₯^2+π¦^2 )ππ₯ ππ/ππ = (π^π + π^π)/(π^π + ππ) Step 2: Putting F(x , y) = ππ¦/ππ₯ and finding F(πx, πy) F(x, y) = (π₯^2 + π¦^2)/(π₯^2 + π₯π¦) Finding F(πx, πy) F(πx, πy) = ((πγπ₯)γ^2+(πγπ¦)γ^2)/((πγπ₯)γ^2+ππ₯ + ππ¦) = (π^2 π₯^2 + π^2 π¦^2)/(π^2 π₯^2 + π^2 π₯π¦) = (π^2 (π₯^2 π¦^2))/(π^2 (π₯^2+ π₯π¦)) = (π₯^2 π¦^2)/(π₯^2 + π₯π¦) = F(x, y) So, F(πx, πy) = F(x, y) = π0 F (x, y) Thus, F(x, y) is a homogenous equation function of order zero Therefore ππ¦/ππ₯ is a homogenous differential equation Step 3 : Solving ππ¦/ππ₯ by putting y = vx Putting y = vx Diff w.r.t.x ππ¦/ππ₯ = xππ£/ππ₯ + v ππ₯/ππ₯ ππ/ππ = x ππ/ππ + v Putting value of ππ¦/ππ₯ and y = vx in (1) ππ¦/ππ₯ = (π₯^2 + π¦^2)/(π₯^2 + π₯π¦) x ππ/ππ + v = (ππ + (ππ)^π)/(ππ + π(ππ)) x ππ£/ππ₯ + v = (ππ(π + π^π))/(π₯2 + π₯2π£) x ππ£/ππ₯ + v = (ππ(1 + π£^2))/(ππ(1 + π£)) x ππ£/ππ₯ + v = (1 + π£^2)/(1 + π£) x ππ£/ππ₯ = (1 + π£^2)/(1 + π£)βπ x ππ£/ππ₯ = (1 + π£^2 β π£ β π£^2)/(1+π£) x ππ/ππ = (π β π)/(π + π) ((1 + π£))/((1 β π£)) dv = ππ₯/π₯ β((π£ + 1)/(π£ β 1)) dv = ππ₯/π₯ ((π + π)/(π β π)) dv = (βππ)/π Integrating both sides β«1β((π£ + 1)/(π£ β 1)) ππ£=ββ«1βππ₯/π₯ β«1β((π + π)/(π β π)) ππ = βlog|π|+π Let I = β«1β((π£ + 1)/(π£ β 1)) ππ£ Solving I I = β«1β((π£ + 1 β 1 + 1)/(π£ β 1)) ππ£ I = β«1β((π β π + π)/(π β π)) ππ£ I = β«1β((π β π )/(π β π)+π/(π β π)) ππ£ I = β«1β(1+2/(π£ β 1)) ππ£ I = β«1βππ£+β«1β2/(π£ β 1) ππ£ I = π+π π₯π¨π β‘γ|πβπ|γ Putting v = y/x I = π¦/π₯+2 logβ‘|π¦/π₯β1| I = π/π+π πππβ‘|(π β π)/π| Putting value of I in (2) π/π+π πππβ‘|(π β π)/π|=βπππβ‘|π|+πͺ π¦/π₯+πππβ‘γ|(π β π)/π|^π γ+logβ‘|π₯|=πΆ π¦/π₯+logβ‘|(π¦ β π₯)^2/π₯^2 |+logβ‘|π₯|=πΆ π¦/π₯+logβ‘|(π β π)^π/π^π Γ π|=πΆ π¦/π₯+logβ‘|(π¦ β π₯)^2/π₯|=πΆ πππβ‘|(π β π)^π/π|=πΆβπ¦/π₯ (π β π)^π/π= π^(πͺ β π/π) (π¦ β π₯)^2/π₯ = π^π Γ π^(β π¦/π₯) (π¦ β π₯)^2/π₯ = π π^(β π¦/π₯) (πβπ)^π = ππ π^(β π/π) (π΄π  π logβ‘π=logβ‘γπ^π γ) (π΄π  logβ‘π+logβ‘π=logβ‘ππ)