Question

Find the equation of the curve passing through the point (1, 1) whose differential equation is x dy = (2 x² + 1) dx (x ≠ 0).

Solution

The given differential equation is
x dy = (2 x² + 1) dx
or

dy space equals space fraction numerator 2 straight x squared plus 1 over denominator straight x end fraction dx

Integrating,

integral space 1 space dy space equals space integral fraction numerator 2 straight x squared plus 1 over denominator straight x end fraction dx

integral 1 space dy space equals space integral open parentheses fraction numerator 2 straight x squared over denominator straight x end fraction plus 1 over straight x close parentheses dx

integral 1 space dy space equals space integral open parentheses 2 straight x plus 1 over straight x close parentheses dx

therefore space space space space space space space space space space space space space space space space space space space space space space space space space space straight y space equals space straight x squared space plus log space open vertical bar straight x close vertical bar space plus straight c

...(1)
The curve passes through (1, 1)

therefore space space space 1 space equals space 1 plus log space open vertical bar 1 close vertical bar plus straight c space space space or space space space space 1 space equals space 1 space plus 0 plus straight c space space space space rightwards double arrow space space space straight c space space equals 0

Putting c = 0 in (1), we get

straight y space equals straight x squared plus space log space open vertical bar straight x close vertical bar comma space which space is space required space solution. space

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