The particular solution of the differential equation xdy = (2x^2 + 1)dx, x ≠ 0, given that y = 1 when x = 1 is: Option 1: y = x + |x| Option 2: y = 2 + |x| Option 3: y = x^2 + |x| Option 4: y = x^2 + |x| + 1

The answer is 'y = x^2 + |x|'. View full solution:

Calculus > Differential Equations

Medium

common

The particular solution of the differential equation $xdy = (2x^2 + 1)dx, x \neq 0$, given that $y = 1$ when $x = 1$ is:

$y = x + \log|x|$

$y = 2 + \log|x|$

$y = x^2 + \log|x|$

$y = x^2 + \log|x| + 1$

✅ Correct Option: 3

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Choose the correct answer from the options given below:

List-I	List-II
(A) Integrating factor of $x \, dy - (y + 2x^2) \, dx = 0$	(I) $\frac{1}{x}$
(B) Integrating factor of $(2x^2 - 3y) \, dx = x \, dy$	(II) $x$
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The particular solution of the differential equation $xdy = (2x^2 + 1)dx, x \neq 0$, given that $y = 1$ when $x = 1$ is:

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