Q15 of 76 Page 929

For each of 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.



Rearranging the terms we get:




Integrating both the sides we get:



⇒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|


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