Examples

Chapter 9 Class 12 Differential Equations
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Example 27 Solve the differential equation (π₯ ππ¦βπ¦ ππ₯)π¦ π ππ(π¦/π₯)=(π¦ ππ₯+π₯ ππ¦)π₯ cosβ‘(π¦/π₯) (π₯ ππ¦βπ¦ ππ₯)π¦ π ππ(π¦/π₯)=(π¦ ππ₯+π₯ ππ¦)π₯ cosβ‘(π¦/π₯) π₯ π¦ sinβ‘γ(π¦/π₯)ππ¦βπ¦^2 π ππ(π¦/π₯) γ ππ₯=π₯π¦ cosβ‘γ(π¦/π₯)ππ₯+π₯^2 γ cosβ‘(π¦/π₯)ππ¦ [π₯ π¦ sinβ‘γ(π¦/π₯)βπ₯^2 πππ (π¦/π₯) γ ]ππ¦=[π₯π¦ cosβ‘γ(π¦/π₯)+π¦^2 γ sinβ‘(π¦/π₯) ]ππ₯ ππ¦/ππ₯ = (π₯π¦ cosβ‘γ(π¦/π₯) + π¦^2 sinβ‘(π¦/π₯) γ)/(π₯π¦ sinβ‘γ(π¦/π₯) β π₯^2 πππ (π¦/π₯)γ ) Dividing numerator & denominator by x2 ππ¦/ππ₯ = (π¦/π₯ cosβ‘γ(π¦/π₯) + (π¦/π₯)^2 cosβ‘(π¦/π₯) γ)/(π¦/π₯ sinβ‘γ(π¦/π₯) β cosβ‘(π¦/π₯) γ ) Putting y = vx. Differentiating w.r.t. x ππ¦/ππ₯ = π₯ ππ£/ππ₯ + v Putting value of ππ¦/ππ₯ and y in (1) ππ¦/ππ₯ = (π¦/π₯ cosβ‘γ(π¦/π₯) + (π¦/π₯)^2 cosβ‘(π¦/π₯) γ)/(π¦/π₯ sinβ‘γ(π¦/π₯) β cosβ‘(π¦/π₯) γ ) v + (π₯ ππ£)/ππ₯=(π£π₯/π₯ cosβ‘γ(π£π₯/π₯) + (π£^2 π₯^2)/π₯^2 sinβ‘(π£π₯/π₯) γ)/(π£π₯/π₯ sinβ‘γ(π£π₯/π₯) β πππ (π£π₯/π₯)γ ) v + (π₯ ππ£)/ππ₯ = (π£ cosβ‘γπ£ + π£^2 sinβ‘π£ γ)/(π£ sinβ‘γπ£ β cosβ‘π£ γ ) x ( ππ£)/ππ₯ = (π£ cosβ‘γπ£ + π£^2 sinβ‘π£ γ)/(π£ sinβ‘γπ£ β cosβ‘π£ γ ) β v x ( ππ£)/ππ₯ = (π£ cosβ‘γπ£ + π£^2 sinβ‘γπ£ β π£(π£ sinβ‘γπ£ β cosβ‘γπ£)γ γ γ γ)/(π£ sinβ‘γπ£ β cosβ‘π£ γ ) x ( ππ£)/ππ₯ = (π£ cosβ‘γπ£ + π£^2 sinβ‘γπ£ β π£^2 sin π£γ + π£ cosβ‘π£ γ)/(π£ sinβ‘γπ£ β cosβ‘π£ γ ) ( π₯ ππ£)/ππ₯ = (2π£ cosβ‘π£)/(π£ sinβ‘γπ£ β cosβ‘π£ γ ) ((π£ sinβ‘γπ£ β cosβ‘π£ γ)/γv cosγβ‘π£ )ππ£=2 ππ₯/π₯ ((π£ sinβ‘π£)/γv cosγβ‘π£ βcosβ‘π£/γv cosγβ‘π£ ) dv = 2 ππ₯/π₯ (tanβ‘π£β1/π£) dv = 2 ππ₯/π₯ Integrating both sides β«1βγ(tanβ‘γπ£ β1/π£γ )ππ£=2β«1βππ₯/π₯γ β«1βtanβ‘γπ£ ππ£ β γ β«1βππ£/π£ = 2 β«1βππ₯/π₯ log |secβ‘π£ |βlogβ‘γ|π£|γ = 2 log |π₯| + log C log |secβ‘π£ |βlogβ‘γ|π£|γ = log |π₯^2 | + log C log |secβ‘π£/π£| = log π₯^2 + log C log |secβ‘π£/π£| = log π₯^2 π secβ‘π£/π£ = π₯^2 π Putting back value of v = π¦/π₯ γsec γβ‘(π¦/π₯)/((π¦/π₯) ) = π₯^2 π secβ‘(π¦/π₯) = (π¦/π₯) π₯^2 π sec (π/π) = C xy (As log a β log b = log π/π & log a + log b = log ab)