Ex 9.4, 10

Transcript

Ex 9.4, 10For each of the differential equations in Exercises 1 to 10, find the general solution : 𝑒^𝑥 tan⁡〖𝑦 𝑑𝑥+(1−𝑒^𝑥 ) sec^2⁡〖𝑦 𝑑𝑦=0〗 〗 𝑒^𝑥 tan⁡〖𝑦 𝑑𝑥+(1−𝑒^𝑥 ) sec^2⁡〖𝑦 𝑑𝑦=0〗 〗 𝑒^𝑥 tan⁡〖𝑦 𝑑𝑥=−(1−𝑒^𝑥 ) sec^2⁡〖𝑦 𝑑𝑦〗 〗 𝑒^𝑥 tan⁡〖𝑦 𝑑𝑥〗=(𝑒^𝑥−1) sec^2⁡〖𝑦 𝑑𝑦〗 𝑒^𝑥/(𝑒^𝑥 − 1) dx = (𝑠𝑒𝑐2𝑦 𝑑𝑦)/tan⁡𝑦 Integrating both sides. ∫1▒〖𝑒^𝑥/(𝑒^𝑥 − 1) 𝑑𝑥〗 = ∫1▒〖(𝑠𝑒𝑐2 𝑦)/tan⁡𝑦 𝑑𝑦〗 Put 𝑒^𝑥−1 = u and put tan y = v Diff u w.r.t. x & v w.r.t y Therefore ∫1▒〖𝑒𝑥/𝑢 𝑑𝑢/𝑒𝑥〗 = ∫1▒𝑠𝑒𝑐2𝑦/(𝑣 𝑠𝑒𝑐2𝑦) dv ∫1▒𝑑𝑢/𝑢 = ∫1▒𝑑𝑣/𝑣 log u + c1 = log v Putting back u = ex − 1 and V = tan y log |"ex − 1" | + c1 = log tan y Putting c1 = log c log |"ex − 1" |+ log c = log (tan y) log |𝑐("ex − 1" )|= log |tan⁡𝑦 | 𝑐("ex − 1" ) = tan y tan y = c ("ex − 1" ) is the general solution