Misc 6 - Chapter 9 Class 12 Differential Equations
Last updated at April 16, 2024 by Teachoo
Variable separation - Equation given
Example 20
Misc 13 (MCQ)
Ex 9.3, 3
Ex 9.3, 8
Example 4
Ex 9.3, 7 Important
Example 5
Ex 9.3, 6
Ex 9.3, 2
Misc 4
Example 6
Ex 9.3, 5
Ex 9.3, 14
Ex 9.3, 13
Example 7 Important
Ex 9.3, 16
Ex 9.3, 12
Ex 9.3, 11 Important
Ex 9.3, 1 Important
Ex 9.3, 15 Important
Misc 7 Important
Ex 9.3, 4 Important
Ex 9.3, 9 Important
Misc 6 You are here
Ex 9.3, 10 Important
Misc 12 Important
Misc 5 Important
Variable separation - Equation given
Last updated at April 16, 2024 by Teachoo
Misc 6 Find the equation of the curve passing through the point (0 , π/4) whose differential equation is sin π₯ cosβ‘γπ¦ ππ₯+cosβ‘γπ₯ sinβ‘γπ¦ ππ¦=0γ γ γ sin x cos y dx + cos x sin y dy = 0 sin x cos y dx = β cos x sin y dy γπ¬π’π§ γβ‘π/ππ¨π¬β‘π π π = β πππβ‘π/πππβ‘π π π Put cos x = u Diff w.r.t. x π/ππ₯ cosβ‘γπ₯=ππ’/ππ₯ γ βsin x = ππ’/ππ₯ dx = (βπ π)/πππβ‘π Put cos y = v Diff w.r.t. y π/ππ¦ cosβ‘γπ¦=ππ£/ππ¦ γ βsin y = ππ£/ππ¦ dy = (βπ π)/πππβ‘π Integrating both sides β«1βsinβ‘γπ₯ ππ₯γ/cosβ‘π₯ = β«1βγβsinγβ‘γπ¦ ππ¦γ/cosβ‘π¦ β΄ β«1βγsinβ‘π₯/π’ Γ (βππ’)/sinβ‘π₯ γ= β«1βγ(βsinβ‘π¦)/π£ Γ (βππ£)/sinβ‘π¦ γ β«1βπ π/π=ββ«1βπ π/π log |π’|=βπππ|π£|+π log |π’| + log |π£| = c log |π’.π£|=π Putting back values of u and v. log |ππ¨π¬β‘γπ.ππ¨π¬β‘π γ |= βc Since the curve passes through (π, π /π) Putting x = 0 and y = π/4 in (1) log |cosβ‘(0).cosβ‘(π/4)|=π log |1. 1/β2|=π β΄ c = log π/βπ Substitute value of C in (2) log |cosβ‘γπ₯ cosβ‘π¦ γ |=π log |ππ¨π¬β‘γπ.ππ¨π¬β‘π γ |=π₯π¨π β‘|π/βπ| β΄ cos x. cos y = π/βπ cos y = 1/(β2 cosβ‘π₯ ) cos y = (πππ π)/βπ