jee-main 2019 Q86

jee-main · India · session2_09apr_shift1 Differential equations First-Order Linear DE: General Solution

The solution of the differential equation $x \frac { d y } { d x } + 2 y = x ^ { 2 } , ( x \neq 0 )$ with $y ( 1 ) = 1$, is
(1) $y = \frac { x ^ { 3 } } { 5 } + \frac { 1 } { 5 x ^ { 2 } }$
(2) $y = \frac { 3 } { 4 } x ^ { 2 } + \frac { 1 } { 4 x ^ { 2 } }$
(3) $y = \frac { x ^ { 2 } } { 4 } + \frac { 3 } { 4 x ^ { 2 } }$
(4) $y = \frac { 4 } { 5 } x ^ { 3 } + \frac { 1 } { 5 x ^ { 2 } }$

The solution of the differential equation $x \frac { d y } { d x } + 2 y = x ^ { 2 } , ( x \neq 0 )$ with $y ( 1 ) = 1$, is\\
(1) $y = \frac { x ^ { 3 } } { 5 } + \frac { 1 } { 5 x ^ { 2 } }$\\
(2) $y = \frac { 3 } { 4 } x ^ { 2 } + \frac { 1 } { 4 x ^ { 2 } }$\\
(3) $y = \frac { x ^ { 2 } } { 4 } + \frac { 3 } { 4 x ^ { 2 } }$\\
(4) $y = \frac { 4 } { 5 } x ^ { 3 } + \frac { 1 } { 5 x ^ { 2 } }$

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