Question

Find the equation of the curve which is passing through the point \( (1, 1) \), and whose differential equation is \[ x \, dy = (2x^2 + 1) \, dx, (x \neq 0). \]

Hint:

To find the equation of a curve from a differential equation, integrate both sides, and use the given point to solve for the constant of integration.

Answer 1

We are given the differential equation: \[ x \, dy = (2x^2 + 1) \, dx. \] First, divide both sides of the equation by \( x \): \[ dy = \frac{2x^2 + 1}{x} \, dx. \] Simplify the right-hand side: \[ dy = (2x + \frac{1}{x}) \, dx. \] Now, integrate both sides: \[ \int dy = \int \left( 2x + \frac{1}{x} \right) \, dx. \] The integral of \( dy \) is \( y \), and the integral of \( 2x \) is \( x^2 \). The integral of \( \frac{1}{x} \) is \( \ln|x| \), so we get: \[ y = x^2 + \ln|x| + C. \]

Step 1: Use the given point to find \( C \). We are given that the curve passes through the point \( (1, 1) \), so substitute \( x = 1 \) and \( y = 1 \) into the equation: \[ 1 = 1^2 + \ln|1| + C. \] Since \( \ln|1| = 0 \), this simplifies to: \[ 1 = 1 + C $\Rightarrow$ C = 0. \] Conclusion: The equation of the curve is: \[ y = x^2 + \ln|x|. \]