# Ex 9.4, 4

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Ex 9.4, 4For 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₯ tanβ‘π¦ ππ₯ +sec^2β‘π¦ tanβ‘γπ₯ ππ¦γ)/tanβ‘γπ¦ tanβ‘π₯ γ = 0/tanβ‘γπ₯ tanβ‘π¦ γ (π ππ2π₯ π‘πππ¦ ππ₯)/tanβ‘γπ¦ tanβ‘π₯ γ + (π ππ2π¦ π‘πππ₯ ππ¦)/tanβ‘γπ¦ tanβ‘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 Therefore 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

Chapter 9 Class 12 Differential Equations

Serial order wise

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