The solution of the differential equation $x{{dy} \over {dx}} + 2y$ = x2 (x $ \ne $ 0) with y(1) = 1, is :

Question

The solution of the differential equation

$x{{dy} \over {dx}} + 2y$ = x² (x $ \ne $ 0) with y(1) = 1, is :

$y = {4 \over 5}{x^3} + {1 \over {5{x^2}}}$

$y = {3 \over 4}{x^2} + {1 \over {4{x^2}}}$

$y = {{{x^2}} \over 4} + {3 \over {4{x^2}}}$

$y = {{{x^3}} \over 5} + {1 \over {5{x^2}}}$

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