Differential Equations Question 49
Question: The solution of the differential equation $ x^{4}\frac{dy}{dx}+x^{3}y+cosec(xy)=0 $ is equal to
[Pb. CET 2004]
Options:
A) $ 2\cos (xy)+{x^{-2}}=c $
B) $ 2\cos (xy)+{y^{-2}}=c $
C) $ 2\sin (xy)+{x^{-2}}=c $
D) $ 2\sin (xy)+{y^{-2}}=c $
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Answer:
Correct Answer: A
Solution:
$ x^{4}\frac{dy}{dx}+x^{3}y+cosec(xy)=0 $
$ x^{4}dy+x^{3}ydx+\text{cosec }(xy)dx=0 $
$ x^{3}(xdy+ydx)+\text{cosec }(xy)dx=0 $
$ x^{3}d(xy)+\text{cosec }(xy)dx=0 $
$ \frac{d(xy)}{\text{cosec }(xy)}+\frac{dx}{x^{3}}=0 $
Integrating both sides, $ \int _{{}}^{{}}{\frac{d(xy)}{\text{cosec }(xy)}}+\int _{{}}^{{}}{\frac{dx}{x^{3}}}=0 $
$ \int _{{}}^{{}}{\sin (xy)d(xy)+\int _{{}}^{{}}{{x^{-3}}dx=0}} $
$ -\cos (xy)+( \frac{{x^{-2}}}{-2} )=c $ ; $ 2\cos (xy)+{x^{-2}}=c $ .
BETA
