Ex 9.4, 8 - Chapter 9 Class 12 Differential Equations
Last updated at Dec. 16, 2024 by Teachoo
Last updated at Dec. 16, 2024 by Teachoo
Ex 9.4, 8 show that the given differential equation is homogeneous and solve each of them. π₯ ππ¦/ππ₯βπ¦+π₯π ππ(π¦/π₯)=0 Step 1: Find ππ¦/ππ₯ π π π/π π = y β x sin (π/π) Step 2: Put ππ¦/ππ₯ = F (x, y) and find F(πx, πy) F(x, y) = π¦/π₯ β sin (π¦/π₯) F(πx, πy) = ("π" π¦)/("π" π₯) β sin (("π" π¦)/("π" π₯)) = π¦/π₯ β sin (π¦/π₯) = F(x, y) = πΒ° [πΉ(π₯, π¦)] β΄ F (x, y) is a homogenous function of degree 0 . So the differential equation ππ¦/ππ₯ is homogenous Step 3: Let y = vx Solving ππ¦/ππ₯= π¦/π₯ - sin (β(π¦@π₯)) Putting y = vx Diff w.r.t.x ππ¦/ππ₯ = x ππ£/ππ₯ + v ππ₯/ππ₯ π π/π π = x π π/π π + v Putting value of ππ¦/ππ₯ = (π₯2 + π¦^2)/(π₯2 + π₯π¦) and y = vx in (1) π₯ ππ¦/ππ₯ = y β x sin (π¦/π₯) v + (π π π)/π π = ππ/π β sin (ππ/π) v + (π₯ ππ£)/ππ₯ = π£ β sin v (π₯ ππ£)/ππ₯ = v β sin v β v (π₯ ππ£)/ππ₯ = βsinβ‘π£ ππ£/ππ₯ = (βsinβ‘π£)/π₯ π π/(πππ π) = (βπ π)/π Integrating both sides β«1βγππ£/(π ππ π£)=β«1β(βππ₯)/π₯γ β«1βγπππ ππ π£ ππ£=ββ«1βππ₯/π₯ γ log |πππππ π βπππβ‘π |=βπππβ‘|π|+πππβ‘π log |πππ ππ π£ βcotβ‘π£ |+logβ‘|π₯|=logβ‘π log |π₯(πππ ππ π£ βcotβ‘γπ£)γ |=logβ‘π x (cosec v β cot v) = C x (1/sinβ‘π£ βcosβ‘π£/sinβ‘π£ ) = C x ((1βcosβ‘γπ£)γ)/sinβ‘π£ = C x(1 β cos v) = C sin v Putting value of v = π¦/π₯ x(πβπππ(π/π)) = C sin (π/π)
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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo