Question

the following differential equations, find a particular solution satisfying the given condition : 

x dy = (2x² + 1) dx (x ≠ 0), given that y = 1 when x = 1.

Answer 1

Rearranging the terms we get:

Integrating both the sides we get:

⇒ \(\int{dy}\) = \(\int{2x\,dx}\) + \(\int\frac{1}{x}dx\) + c

⇒ y = x² + log|x| + c 

y = 1 when x = 1 

∴1 = 1² + log1 + c 

∴1 - 1 = 0 + c …(log1 = 0) 

⇒ c = 0 

∴y = x² + log|x| 

Ans: y = x² + log |x|

Answer 2

The solution is provided in the image below. 

Both LHS and RHS are divided by x later to which integration is done on both sides to obtain a general solution. The given condition is put in the general solution to obtain a particular solution.