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Differential Equations question

2019 · 9 Apr · Shift 1 · Q41

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JEE Main
Mathematics
Differential Equations
2019 · 9 Apr · Shift 1 · Q41

Differential Equations question

2019 · 9 Apr · Shift 1 · Q41

JEE MainMathematicsDifferential EquationsMCQ+4 / −1

The solution of the differential equation

x{{dy} \over {dx}} + 2y

= x2 (x

e

0) with y(1) = 1, is :

A
$y = {4 \over 5}{x^3} + {1 \over {5{x^2}}}$
B
$y = {3 \over 4}{x^2} + {1 \over {4{x^2}}}$
C
$y = {{{x^2}} \over 4} + {3 \over {4{x^2}}}$
D
$y = {{{x^3}} \over 5} + {1 \over {5{x^2}}}$

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