A) \[\frac{7}{64}\]
B) \[\frac{49}{16}\]
C) \[\frac{1}{4}\]
D) \[\frac{13}{16}\]
Correct Answer: B
Solution :
\[x\frac{dy}{dx}\,\,+\,\,2y\,\,=\,\,{{x}^{2}}\,\] \[\frac{dy}{dx}\,\,+\,\,\frac{2y}{x}\,\,=\,\,x\,\] \[I.F.\,\,=\,\,{{e}^{\int{\frac{2}{x}.\,dx\,\,}}}\,=\,\,{{e}^{2\ln x\,\,=\,{{x}^{2}}}}\] \[y.{{x}^{2}}\,=\,\int{x.{{x}^{2}}\,.\,dx\,\,+\,\,c}\] \[y{{x}^{2}}\,\,=\,\,\frac{{{x}^{4}}}{4}\,\,+\,\,c\] \[y\left( 1 \right)\,\,=\text{ }1\] \[1.\,\,1=\frac{1}{4}\,+\,c\,\,\,\Rightarrow \,c=1-\frac{1}{4}\,\,\Rightarrow \,c=\frac{3}{4}\] \[y{{x}^{2}}\,\,=\,\,\frac{{{x}^{4}}}{4}\,\,+\,\,\frac{3}{4}\] \[y.\,\,\frac{1}{4}\,\,=\,\,\frac{1}{4}\,.\,\frac{1}{16}\,\,+\,\,\frac{3}{4}\] \[y\,\,=\,\,\frac{1}{16}\,\,+\,\,3\] \[y\,\,=\,\,\frac{49}{16}\]
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