Let x = x(y) be the solution of the differential equation $$2y\,{e^{x/{y^2}}}dx + \left( {{y^2} - 4x{e^{x/{y^2}}}} \right)dy = 0$$ such that x(1) = 0. Then, x(e) is equal to :
JEE · Math · previous-year question
- A.$$e{\log _e}(2)$$
- B.$$ - e{\log _e}(2)$$
- C.$${e^2}{\log _e}(2)$$
- D.$$ - {e^2}{\log _e}(2)$$correct
Answer
D. $$ - {e^2}{\log _e}(2)$$
Explanation
Given differential equation $$2y{e^{{x \over {{y^2}}}}}dx + \left( {{y^2} - 4x{e^{{x \over {{y^2}}}}}} \right)dy = 0,\,x(1) = 0$$ $$ \Rightarrow {e^{{x \over {{y^2}}}}}[2ydx - 4xdy] = - {y^2}dy$$ $$ \Rightarrow {e^{{x \over {{y^2}}}}}\left[ {{{2{y^2}dx - 4xydy} \over {{y^4}}}} \right] = {{ - 1} \over y}dy$$ $$ \Rightarrow 2{e^{{x \over {{y^2}}}}}d\left( {{x \over {{y^2}}}} \right) = - {1 \over y}dy$$ $$ \Rightarrow 2{e^{{x \over {{y^2}}}}} = - \ln y + c$$ ...... (i) Now, using x(1) = 0, c = 2 So, for x(e), Put y = e in (i) $$2{e^{{x \over {{e^2}}}}} = - 1 + 2$$ $$ \Rightarrow {x \over {{e^2}}} = \ln \left( {{1 \over 2}} \right) \Rightarrow x(e) = - {e^2}\ln 2$$
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