Official solution
\[x{{dy} \over {dx}} + 2y\] = x2
\[ \Rightarrow \] \[{{dy} \over {dx}} + \left( {{2 \over x}} \right)y = x\]
\[ \therefore \] I.F = \[{e^{\int {{2 \over x}dx} }}\] = x2
\[ \therefore \] The solution is
yx2 = \[\int {{x^3}dx} \]
\[ \Rightarrow \] yx2 = \[{{{x^4}} \over 4} + C\] .....(1)
As y(1) = 1
\[ \therefore \] when x = 1 then y = 1.
Putting the value of x and y in equation (1), we get
1 = \[{1 \over 4} + C\]
\[ \Rightarrow \] C = \[{3 \over 4}\]
\[ \therefore \] Required solution
yx2 = \[{{{x^4}} \over 4} + {3 \over 4}\]
\[ \Rightarrow \] \[y = {{{x^2}} \over 4} + {3 \over {4{x^2}}}\]