The solution of the differential equation x dy dx + y = x^2 + 3x … - Vidyadip

MCQ

The solution of the differential equation $x\frac{{dy}}{{dx}} + y = {x^2} + 3x + 2$ is

✓
$xy = \frac{{{x^3}}}{3} + \frac{3}{2}{x^2} + 2x + c$
B
$xy = \frac{{{x^4}}}{4} + {x^3} + {x^2} + c$
C
$xy = \frac{{{x^4}}}{4} + \frac{{{x^3}}}{3} + {x^2} + c$
D
$xy = \frac{{{x^4}}}{4} + {x^3} + {x^2} + cx$

✓

Answer

Correct option: A.

$xy = \frac{{{x^3}}}{3} + \frac{3}{2}{x^2} + 2x + c$

a
(a) $x\frac{{dy}}{{dx}} + y = {x^2} + 3x + 2$ ==> $\frac{{dy}}{{dx}} + \frac{y}{x} = x + 3 + \frac{2}{x}$

Here$P = \frac{1}{x},\;Q = x + 3 + \frac{2}{x}$, therefore $I.F.$ $_{e}\int_{{}}^{{}}{\frac{1}{x}}dx=x$

Now solve it.

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