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Find the general solution of the differential equations `e^xtanydx+(1-e^x)sec^2ydy=0` |
Answer» `e^(x)tanydx+(1-e^(x))sec^(2)ydy=0` `implies e^(x)tany(dx=-(1-e^(x))sec^(2)ydy` `implies (e^(x))/((e^(x)-1))dx=(sec^(2))/(tany)dy` `implies int(e^(x))/((e^(x)-1))dx=int(sec^(2))/(tany)dy` Let `e^(x)=1-timplies e^(x)dx=dt` Let `tany=vimplies sec^(2)ydy=dv` `:. int(1)/(t)dt=int(1)/(v)dv` `implies log|t|=log|v|-log|C|` `implies log|e^(x)-1|=log|tany|-log|C|` `implies log|C(e^(x)-1)|=log|tany|` `implies C(e^(x)-1)=tany` which is the required general solution. |
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