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Question
Find the equation of the curve passing through the point (1, 1) whose differential equation is x dy = (2x2 + 1) dx, x ≠ 0.
Solution
We have,
\[x dy = \left( 2 x^2 + 1 \right)dx\]
\[ \Rightarrow dy = \left( \frac{2 x^2 + 1}{x} \right)dx\]
\[ \Rightarrow dy = \left( 2x + \frac{1}{x} \right)dx\]
Integrating both sides, we get
\[\int dy = \int\left( 2x + \frac{1}{x} \right)dx\]
\[ \Rightarrow y = x^2 + \log \left| x \right| + C . . . . . . . . . . \left( 1 \right)\]
Now the given curve passes through (1, 1)
Therefore, when x = 1, y = 1\]
\[ \therefore 1 = 1 + 0 + C\]
\[ \Rightarrow C = 0\]
Putting the value of `C` in (1), we get
\[y = x^2 + \log\left| x \right|\]
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