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The solution of `x^(3)(dy)/(dx)+4x^(2)tan y=e^(x)sec y` satisfying y(1) = 0, isA. `tan y = e^(x)(x-2)lnx`B. `siny=e^(x)(x-1)x^(-4)`C. `tany=e^(x)(x-1)x^(-3)`D. `siny=e^(x)(x-1)x^(-3)` |
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Answer» Correct Answer - B `x^(3)(dy)/(dx)+4x^(2)tany=e^(x)secyimpliesx^(3)cosy(dy)/(dx)+4x^(2)siny =e^(x)` `implies x^(3).(dt)/(dx)+4x^(2).t=e^(x), " "("where " t=siny)` `impliesx^(4)(dt)/(dx)+4x^(3)t=xe^(x)implies(d)/(dx)(x^(4).t)=xe^(x)impliesx^(4)t=(x-1)e^(x)+c` `impliesx^(4)siny=(x-1)e^(x)+c` `becausey(1)=0impliesc=0impliessiny=(e^(x)(x-1))/(x^(4))` |
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