Ex 9.4

Chapter 9 Class 12 Differential Equations
Serial order wise

### Transcript

Ex 9.4, 8 show that the given differential equation is homogeneous and solve each of them. π₯ ππ¦/ππ₯βπ¦+π₯π ππ(π¦/π₯)=0 Step 1: Find ππ¦/ππ₯ π ππ/ππ = y β x sin (π/π) Step 2: Put ππ¦/ππ₯ = F (x, y) and find F(πx, πy) F(x, y) = π¦/π₯ β sin (π¦/π₯) F(πx, πy) = ("π" π¦)/("π" π₯) β sin (("π" π¦)/("π" π₯)) = π¦/π₯ β sin (π¦/π₯) = F(x, y) = πΒ° [πΉ(π₯, π¦)] β΄ F (x, y) is a homogenous function of degree 0 . So the differential equation ππ¦/ππ₯ is homogenous Step 3: Let y = vx Solving ππ¦/ππ₯= π¦/π₯ - sin (β(π¦@π₯)) Putting y = vx Diff w.r.t.x ππ¦/ππ₯ = x π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₯ = (π₯2 + π¦^2)/(π₯2 + π₯π¦) and y = vx in (1) π₯ ππ¦/ππ₯ = y β x sin (π¦/π₯) v + (π ππ)/ππ = ππ/π β sin (ππ/π) v + (π₯ ππ£)/ππ₯ = π£ β sin v (π₯ ππ£)/ππ₯ = v β sin v β v (π₯ ππ£)/ππ₯ = βsinβ‘π£ ππ£/ππ₯ = (βsinβ‘π£)/π₯ ππ/(πππ π) = (βππ)/π Integrating both sides β«1βγππ£/(π ππ π£)=β«1β(βππ₯)/π₯γ β«1βγπππ ππ π£ ππ£=ββ«1βππ₯/π₯ γ log |πππππ π βπππβ‘π |=βπππβ‘|π|+πππβ‘π log |πππ ππ π£ βcotβ‘π£ |+logβ‘|π₯|=logβ‘π log |π₯(πππ ππ π£ βcotβ‘γπ£)γ |=logβ‘π x (cosec v β cot v) = C x (1/sinβ‘π£ βcosβ‘π£/sinβ‘π£ ) = C x ((1βcosβ‘γπ£)γ)/sinβ‘π£ = C x(1 β cos v) = C sin v Putting value of v = π¦/π₯ x(πβπππ(π/π)) = C sin (π/π)