Ex 9.3

Chapter 9 Class 12 Differential Equations
Serial order wise

### Transcript

Ex 9.3, 4 For each of the differential equations in Exercises 1 to 10, find the general solution : sec^2β‘γπ₯ tanβ‘γπ¦ ππ₯+sec^2β‘γπ¦ tanβ‘γπ₯ππ¦=0γ γ γ γsec^2β‘γπ₯ tanβ‘γπ¦ ππ₯+sec^2β‘γπ¦ tanβ‘γπ₯ππ¦=0γ γ γ γ Dividing both sides by tan y tan x (π ππ2π₯ π‘ππβ‘π¦ ππ₯ +γπ ππγ^2β‘π¦ π‘ππβ‘γπ₯ ππ¦γ)/π‘ππβ‘γπ¦ π‘ππβ‘π₯ γ = 0/π‘ππβ‘γπ₯ π‘ππβ‘π¦ γ (π ππ2π₯ π‘πππ¦ ππ₯)/π‘ππβ‘γπ¦ π‘ππβ‘π₯ γ + (π ππ2π¦ π‘πππ₯ ππ¦)/π‘ππβ‘γπ¦ π‘ππβ‘π₯ γ = 0 (πππππ )/π­ππ§β‘π dx + πππππ/πππβ‘π dy = 0 Integrating both sides β«1βγ(π ππ2π₯/tanβ‘π₯ ππ₯+π ππ2π¦/tanβ‘π¦ ππ¦)=γ 0 β«1βγπππππ/πππβ‘π ππ+β«1βγπππππ/πππβ‘π ππγ=γ 0 Put u = tan x and v = tan y Diff u w.r.t. x & v w.r.t y Therefore, our equation becomes β«1βγsec^2β‘π₯/tanβ‘π₯ ππ₯γ+β«1βγsec^2β‘π¦/tanβ‘π¦ ππ¦=0γ ππ’/ππ₯ = π ππ2π₯ ππ/πππππ" " = dx ππ£/ππ¦ = π ππ2π¦ ππ/πππππ" " = dy β«1βγsec^2β‘π₯/tanβ‘π₯ ππ’/sec^2β‘π₯ γ+β«1βγ sec^2β‘π¦/π£ ππ£/sec^2β‘π¦ =0γ β«1βγππ’/π’+β«1βγππ£/π£=0γγ log |π|+πππβ‘|π|=πππβ‘π Putting back u = tan x and v = tan y log |πππβ‘π | + log |π­ππ§β‘π |=π₯π¨π β‘π log |tanβ‘γπ₯+tanβ‘π¦ γ | =logβ‘π tan x tan y = C