InterviewSolution
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InterviewSolution
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`e^(x)tanydx+(1-e^(x))sec^(2)ydy=0` |
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Answer» `e^(x)tanydx+(1-e^(x))sec^(2)ydy=0` `implies e^(x)tan y dx=-(1-e^(x))sec^(2)y dy` `implies (e^(x))/((e^(x)-1))dx=(sec^(2)y)/(tan y)dy` `implies int(e^(x))/((e^(x)-1))dx=int(sec^(2)y)/(tany)dy` माना `e^(x)-1=t implies e^(x)dx=dt` माना `tan y = v implies sec^(2)y dy = dv` `therefore int (1)/(t)dt = int(1)/(v) dv` `implies log|t|=log|v|-log|C|` `implies log|e^(x)-1|=log|tany|-log|C|` `log|C(e^(x)-1)|=log|tany|` `implies C^(e^(x)-1)=tan y` जोकि अभीष्ट व्यापक हल है। |
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