Last updated at Feb. 17, 2025 by Teachoo
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 )πx (π₯^2+π₯π¦)ππ¦ =(π₯^2+π¦^2 )ππ₯ Step 1: Find ππ¦/ππ₯ (x2 + xy)dy = (π₯^2+π¦^2 )ππ₯ π π/π π = (π^π + π^π)/(π^π + ππ) Step 2: Putting F(x , y) = ππ¦/ππ₯ and finding F(πx, πy) F(x, y) = (π₯^2 + π¦^2)/(π₯^2 + π₯π¦) Finding F(πx, πy) F(πx, πy) = ((πγπ₯)γ^2+(πγπ¦)γ^2)/((πγπ₯)γ^2+ππ₯ + ππ¦) = (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(πx, πy) = F(x, y) = π0 F (x, y) Thus, F(x, y) is a homogenous equation function of order zero Therefore ππ¦/ππ₯ is a homogenous differential equation Step 3 : Solving ππ¦/ππ₯ by putting y = vx Putting y = vx Diff w.r.t.x ππ¦/ππ₯ = xππ£/ππ₯ + v ππ₯/ππ₯ π π/π π = x π π/π π + v Putting value of ππ¦/ππ₯ and y = vx in (1) ππ¦/ππ₯ = (π₯^2 + π¦^2)/(π₯^2 + π₯π¦) x π π/π π + v = (ππ + (ππ)^π)/(ππ + π(ππ)) x ππ£/ππ₯ + v = (ππ(π + π^π))/(π₯2 + π₯2π£) x ππ£/ππ₯ + v = (ππ(1 + π£^2))/(ππ(1 + π£)) x ππ£/ππ₯ + v = (1 + π£^2)/(1 + π£) x ππ£/ππ₯ = (1 + π£^2)/(1 + π£)βπ x ππ£/ππ₯ = (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 = ππ₯/π₯ β((π£ + 1)/(π£ β 1)) dv = ππ₯/π₯ ((π + π)/(π β π)) dv = (βπ π)/π Integrating both sides β«1β((π£ + 1)/(π£ β 1)) ππ£=ββ«1βππ₯/π₯ β«1β((π + π)/(π β π)) π π = βlog|π|+π Let I = β«1β((π£ + 1)/(π£ β 1)) ππ£ Solving I I = β«1β((π£ + 1 β 1 + 1)/(π£ β 1)) ππ£ I = β«1β((π β π + π)/(π β π)) ππ£ I = β«1β((π β π )/(π β π)+π/(π β π)) ππ£ I = β«1β(1+2/(π£ β 1)) ππ£ I = β«1βππ£+β«1β2/(π£ β 1) ππ£ I = π+π π₯π¨π β‘γ|πβπ|γ Putting v = y/x I = π¦/π₯+2 logβ‘|π¦/π₯β1| I = π/π+π πππβ‘|(π β π)/π| Putting value of I in (2) π/π+π πππβ‘|(π β π)/π|=βπππβ‘|π|+πͺ π¦/π₯+πππβ‘γ|(π β π)/π|^π γ+logβ‘|π₯|=πΆ π¦/π₯+logβ‘|(π¦ β π₯)^2/π₯^2 |+logβ‘|π₯|=πΆ π¦/π₯+logβ‘|(π β π)^π/π^π Γ π|=πΆ π¦/π₯+logβ‘|(π¦ β π₯)^2/π₯|=πΆ πππβ‘|(π β π)^π/π|=πΆβπ¦/π₯ (π β π)^π/π= π^(πͺ 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₯) (πβπ)^π = ππ π^(β π/π) (π΄π π logβ‘π=logβ‘γπ^π γ) (π΄π logβ‘π+logβ‘π=logβ‘ππ)
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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 15 years. He provides courses for Maths, Science and Computer Science at Teachoo