Ex 9.4

Chapter 9 Class 12 Differential Equations
Serial order wise

### Transcript

Ex 9.4, 7 Show that the given differential equation is homogeneous and solve each of them. {π₯πππ (π¦/π₯)+π¦ sinβ‘(π¦/π₯) }π¦ ππ₯={π¦π ππ(π¦/π₯)βπ₯ cosβ‘(π¦/π₯) }π₯ ππ¦ Step 1: Find ππ¦/ππ₯ {π₯πππ (π¦/π₯)+π¦ sinβ‘(π¦/π₯) }π¦ ππ₯={π¦π ππ(π¦/π₯)βπ₯ cosβ‘(π¦/π₯) }π₯ ππ¦ ππ/ππ=((π πππβ‘(π/π) + π πππβ‘(π/π))/(π πππβ‘(π/π) β π πππβ‘γ (π/π)γ ))" " π/π Step 2: Put ππ¦/ππ₯ = F (x, y) and find F(πx, πy) F(x, y) = ((π₯ cosβ‘(π¦/π₯) + π¦ sinβ‘(π¦/π₯))/(π¦ sinβ‘(π¦/π₯) β π₯ cosβ‘γ (π¦/π₯)γ )) π¦/π₯ F(πx, πy) = ((ππ₯ cosβ‘( ππ¦/ππ₯) + ππ¦ sinβ‘( ππ¦/ππ₯ ))/(ππ₯ sinβ‘( ππ¦/ππ₯) β ππ₯ cosβ‘γ ( ππ¦/ππ₯ )γ )) ππ¦/ππ₯ = ((π₯ cosβ‘(π¦/π₯) + π¦ sinβ‘(π¦/π₯))/(π¦ sinβ‘(π¦/π₯) β π₯ cosβ‘γ (π¦/π₯)γ ))" " π¦/π₯ = F(x, y) β΄ F(πx, πy) = F(x, y) = πΒ° F(x, y) Hence, F(x, y) is a homogenous Function of with degree zero So, ππ¦/ππ₯ is a homogenous differential equation. Step 3: Solving ππ¦/ππ₯ by putting y = vx Putting y = vx. Differentiating w.r.t.x ππ¦/ππ₯ = x ππ£/ππ₯+π£ππ₯/ππ₯ ππ/ππ = π ππ/ππ + v Putting value of ππ¦/ππ₯ and y = vx in (1) ππ¦/ππ₯=((π₯ cosβ‘(π¦/π₯) + π¦ sinβ‘(π¦/π₯))/(π¦ sinβ‘(π¦/π₯) β π₯ cosβ‘γ (π¦/π₯)γ ))" " π¦/π₯ v + x ππ/ππ = (π πππβ‘(ππ/π) + ππ πππβ‘(ππ/π))/(ππ πππβ‘(ππ/π) β π πππβ‘(ππ/π) ) Γ ππ/π v + x ππ£/ππ₯ = [π₯πππ  π£ + π£π₯ π ππ π£]π£/(π£π₯ sinβ‘γπ£ β π₯ cosβ‘π£ γ ) v + x ππ£/ππ₯ = π₯[πππ  π£ + π£ π ππ π£]π£/π₯[π£ π ππ π£ βγ cosγβ‘π£ ] v + x ππ£/ππ₯ = π£[πππ  π£ + π£ π ππ π£]/(π£ π ππ π£ βγ cosγβ‘π£ ) x ( ππ£)/ππ₯ = (π£ cosβ‘π£ + π£^2 sinβ‘π£)/(π£ π ππ π£ βγ cosγβ‘π£ ) β v x ( ππ£)/ππ₯ = (π£ cosβ‘π£ + π£^2 sinβ‘π£ β π£(π£ sinβ‘γπ£ γβ cosβ‘π£ ))/(π£ π ππ π£ βγ cosγβ‘π£ ) x ( ππ£)/ππ₯ = (π£ cosβ‘π£ + π£^2 sinβ‘π£ β π£^2 sinβ‘γπ£ + π£ cosβ‘π£ γ)/(π£ π ππ π£ βγ cosγβ‘π£ ) ( π₯ππ£)/ππ₯ = (2π£ cosβ‘π£)/(π£ π ππ π£ βγ cosγβ‘π£ ) ( ππ)/ππ = ( π)/π [(ππ πππβ‘π)/(π πππ π βγ πππγβ‘π )] (π£ sinβ‘γπ£ β cosβ‘π£ γ)/(π£ γ cosγβ‘π£ ) dv = 2 ( ππ₯)/π₯ Integrating both sides. β«1β(π πππβ‘γπ βπππβ‘π γ)/(π γ πππγβ‘π ) dv = 2 ( ππ)/π β«1β(π£ sinβ‘γπ£ γ)/(π£ γ cosγβ‘π£ ) dv β β«1βcosβ‘π£/(π£ γ cosγβ‘π£ ) dv = 2β«1βππ₯/π₯ β«1βsinβ‘γπ£ γ/cosβ‘π£ dv β β«1βππ£/π£ dv = 2β«1βππ₯/π₯ β«1βtanβ‘π£ dv β β«1βππ£/π£ dv = 2β«1βππ₯/π₯ logβ‘γ|secβ‘π₯ |βlogβ‘|π£|=2 logβ‘|π₯| γ + C πππβ‘γ|πππβ‘π/π|=πππβ‘γπ^π γ+π γ logβ‘γ|secβ‘π₯/π£|=logβ‘γπ₯^2 γ+logβ‘π γ logβ‘γ|secβ‘π₯/π£|=logβ‘γππ₯^2 γ γ Put v = π¦/π₯ log |π¬ππβ‘(π/π)/(π/π)| = log (cx2) |γsec γβ‘γ(π¦/π₯) γ/(π¦/π₯)| = cx2 |γsec γβ‘γ(π¦/π₯) γ | = cx2 Γ π¦/π₯ |γπππ γβ‘(π/π) |= C 1/|cosβ‘(π¦/π₯) | = C xy xy |πππ (π¦/π₯)| = c1 xy πππ|π/π| = c1