A) \[xy={{x}^{4}}+c\]
B) \[xy={{y}^{4}}+c\]
C) \[4xy={{y}^{4}}+c\]
D) \[3xy={{y}^{3}}+c\] where c is an arbitrary constant.
Correct Answer: C
Solution :
[c] \[\frac{dx}{dy}+\frac{x}{y}-{{y}^{2}}=0;\frac{dx}{dy}+\frac{x}{y}={{y}^{2}}\] This is a linear differential equation of the form \[\frac{dx}{dy}+{{P}_{1}}x={{Q}_{1}};\] Here, \[P=\frac{1}{y}\] and \[Q={{y}^{2}}\] \[\therefore \] I.F. \[={{e}^{\int{Pdy}}}={{e}^{\int{\frac{1}{y}dy}}}={{e}^{\log y}}=y\] So, required solution is \[x\cdot y=\int{{{y}^{2}}\cdot ydy+c;xy=\int{{{y}^{3}}dy+c}}\] \[xy=\frac{{{y}^{4}}}{4}+c;4xy={{y}^{4}}+c\]
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