A) \[xy=\frac{1}{3}{{y}^{3}}+c\]
B) \[xy={{y}^{4}}+c\]
C) \[{{y}^{4}}=4xy+c\]
D) \[4y={{y}^{3}}+c\]
Correct Answer: C
Solution :
\[ydx+(x-{{y}^{3}})dy=0\] Here, \[M=y\] and \[N=x-{{y}^{3}}.\]We have, \[\frac{\partial M}{\partial y}=1\]and \[\frac{\partial N}{\partial x}=1,\]\[i.e.,\]\[\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}\] Hence, the given equation is exact. Integrating M, i.e., \[y\] w.r.t. \[x,\]treating \[y\] as a constant, we get\[xy\]. Again, the only terms in N which do not contain \[x\] are\[-{{y}^{3}}.\] Integrating\[(-{{y}^{3}})\]w.r.t. \[y\]we obtain\[-\frac{{{y}^{4}}}{4}.\] Hence, the solution of the given differential equation is \[xy-\frac{{{y}^{4}}}{4}+c=0\] \[\Rightarrow \] \[{{y}^{4}}=4xy+c\]
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