Solve the differential equation dy/dx = 1/(xy) + y/x given that y = √3 when x = 1 - Leaving Cert Applied Maths - Question 10 - 2009

To solve the differential equation, we start from:

$\frac{dy}{dx} = \frac{1}{xy} + \frac{y}{x}$

Rearranging gives us:

$dy = \left(\frac{1}{xy} + \frac{y}{x}\right) dx$

Integrating both sides:

$\int y + y^2 dy = \int \left(\frac{1}{x} + \frac{1}{x} y\right) dx$

This results in:

$\frac{1}{2} \ln(1+y^2) = \ln x + C$

Exponentiating to solve for y, we substitute the initial condition y = √3 when x = 1:

After performing the integration and back-substituting, we find a specific solution for y.