

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
Ex 9.3
Ex 9.3, 2
Ex 9.3, 3
Ex 9.3, 4 Important You are here
Ex 9.3, 5
Ex 9.3, 6
Ex 9.3, 7 Important
Ex 9.3, 8
Ex 9.3, 9 Important
Ex 9.3, 10 Important
Ex 9.3, 11 Important
Ex 9.3, 12
Ex 9.3, 13
Ex 9.3, 14
Ex 9.3, 15 Important
Ex 9.3, 16
Ex 9.3, 17 Important
Ex 9.3, 18
Ex 9.3, 19 Important
Ex 9.3, 20 Important
Ex 9.3, 21
Ex 9.3, 22 Important
Ex 9.3, 23 (MCQ)
Last updated at May 29, 2023 by Teachoo
Ex 9.3, 4 For each of the differential equations in Exercises 1 to 10, find the general solution : sec^2β‘γπ₯ tanβ‘γπ¦ ππ₯+sec^2β‘γπ¦ tanβ‘γπ₯ππ¦=0γ γ γ γ sec^2β‘γπ₯ tanβ‘γπ¦ ππ₯+sec^2β‘γπ¦ tanβ‘γπ₯ππ¦=0γ γ γ γ Dividing both sides by tan y tan x (π ππ2π₯ π‘ππβ‘π¦ ππ₯ +γπ ππγ^2β‘π¦ π‘ππβ‘γπ₯ ππ¦γ)/π‘ππβ‘γπ¦ π‘ππβ‘π₯ γ = 0/π‘ππβ‘γπ₯ π‘ππβ‘π¦ γ (π ππ2π₯ π‘πππ¦ ππ₯)/π‘ππβ‘γπ¦ π‘ππβ‘π₯ γ + (π ππ2π¦ π‘πππ₯ ππ¦)/π‘ππβ‘γπ¦ π‘ππβ‘π₯ γ = 0 (π ππ2π₯ )/tanβ‘π₯ dx + π ππ2π¦/tanβ‘π¦ dy = 0 Integrating both sides β«1βγ(π ππ2π₯/tanβ‘π₯ ππ₯+π ππ2π¦/tanβ‘π¦ ππ¦)=γ 0 β«1βγπ ππ2π₯/tanβ‘π₯ ππ₯+β«1βγπ ππ2π¦/tanβ‘π¦ ππ¦γ=γ 0 Put u = tan x and v = tan y Diff u w.r.t. x & v w.r.t y ππ’/ππ₯ = π ππ2π₯ ππ’/π ππ2π₯" " = dx ππ£/ππ¦ = π ππ2π¦ ππ’/π ππ2π₯" " = dy Therefore, our equation becomes β«1βγsec^2β‘π₯/tanβ‘π₯ ππ₯γ+β«1βγsec^2β‘π¦/tanβ‘π¦ ππ¦=0γ β«1βγsec^2β‘π₯/tanβ‘π₯ ππ’/sec^2β‘π₯ γ+β«1βγ sec^2β‘π¦/π£ ππ£/sec^2β‘π¦ =0γ β«1βγππ’/π’+β«1βγππ£/π£=0γγ log |π’|+logβ‘|π£|=logβ‘π Putting back u = tan x and v = tan y log |tanβ‘π₯ | + log |tanβ‘π¦ |=logβ‘π log |tanβ‘γπ₯+tanβ‘π¦ γ | =logβ‘π tan x tan y = C (β΅logβ‘γπ+logβ‘γπ=logβ‘ππ γ γ )