A) \[{{x}^{2}}y+1=3x\]
B) \[{{x}^{2}}y+1=0\]
C) \[xy+1=3x\]
D) \[{{x}^{2}}(y+1)=3x\]
E) \[{{x}^{2}}y=3x+1\]
Correct Answer: A
Solution :
\[x\frac{dy}{dx}+y=\frac{1}{{{x}^{2}}}\] \[\Rightarrow \] \[\frac{dy}{dx}+\frac{y}{x}=\frac{1}{{{x}^{3}}}\] \[IF={{e}^{\int{\frac{dx}{x}}}}={{e}^{\log x}}=x\] Complete solution, \[y.(IF)=\int{\frac{1}{{{x}^{3}}}}.(IF)dx+C\] \[yx=\int{\frac{1}{{{x}^{3}}}}.x\,dx+C\] \[yx=\int{\frac{dx}{{{x}^{2}}}}+C\] \[yx=\frac{-1}{x}+C\] \[\Rightarrow \] \[y=\frac{-1}{{{x}^{2}}}+\frac{C}{x}\] At \[(1,\text{ }2)\] \[2=-1+C\] \[\Rightarrow \] \[C=3\] Hence, the required solution is, \[y=-\frac{1}{{{x}^{2}}}+\frac{3}{x}\] Or \[{{x}^{2}}y+1=3x\]
You need to login to perform this action.
You will be redirected in
3 sec
