Question

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

Answer 1

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

\[y = x^2 + \log\left| x \right|\]