# Misc 6 - Chapter 9 Class 12 Differential Equations

Last updated at April 16, 2024 by Teachoo

Miscellaneous

Misc 1 (i)

Misc 1 (ii)

Misc 1 (iii) Important

Misc 2 (i)

Misc 2 (ii) Important

Misc 2 (iii)

Misc 2 (iv) Important

Misc 3 Important

Misc 4

Misc 5 Important

Misc 6 You are here

Misc 7 Important

Misc 8 Important

Misc 9

Misc 10 Important

Misc 11

Misc 12 Important

Misc 13 (MCQ)

Misc 14 (MCQ) Important

Misc 15 (MCQ)

Question 1 Deleted for CBSE Board 2025 Exams

Question 2 Important Deleted for CBSE Board 2025 Exams

Question 3 Important Deleted for CBSE Board 2025 Exams

Chapter 9 Class 12 Differential Equations

Serial order wise

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