A) \[3{{x}^{2}}{{y}^{3}}+4{{x}^{3}}=c{{y}^{3}}\]
B) \[3{{x}^{4}}{{y}^{3}}+4{{x}^{3}}=c{{y}^{3}}\]
C) \[3{{x}^{6}}{{y}^{3}}+4{{x}^{3}}=c{{y}^{3}}\]
D) \[3{{x}^{4}}{{y}^{3}}-4{{x}^{3}}=c{{y}^{3}}\]
Correct Answer: B
Solution :
\[\frac{dy}{dx}-\frac{y}{x}={{y}^{4}}\] \[\Rightarrow \] \[\frac{dy}{dx}-\frac{y}{x}={{y}^{4}}\] \[\Rightarrow \] \[{{y}^{-4}}\frac{dy}{dx}-\frac{{{y}^{-3}}}{x}=1\] \[{{y}^{-3}}=t\] \[\Rightarrow \] \[-3{{y}^{-4}}\frac{dy}{dx}=\frac{dt}{dx}\] \[\Rightarrow \] \[-\frac{1}{3}\frac{dt}{dx}-\frac{t}{x}=1\] \[\Rightarrow \] \[\frac{dt}{dx}+\frac{3t}{x}=-3\] \[I.F.={{e}^{3\ell nx}}={{x}^{3}}\] Solution is given by \[t\cdot {{x}^{3}}=-\frac{3{{x}^{4}}}{4}+c\] \[\Rightarrow \] \[4{{x}^{3}}=-3{{x}^{4}}{{y}^{3}}+4c{{y}^{3}}\] \[\Rightarrow \] \[3{{x}^{4}}{{y}^{3}}+4{{x}^{3}}=k{{y}^{3}}\]
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