**The question:**
Use the substitution $\displaystyle v = \frac{y}{x}$ to solve

$\displaystyle \frac{dy}{dx} = \frac{(\frac{y}{x})^3 + 1}{(\frac{y}{x})^2}$

given that y = 1 at x = 1.

**My attempt:**
$\displaystyle \frac{dy}{dx} = \frac{v^3 + 1}{v^2}$

$\displaystyle y = \int{\frac{v^3}{v^2}} + \int{\frac{1}{v^2}}$

= $\displaystyle \int{v} + \int{\frac{1}{v^2}}$

= $\displaystyle \frac{v^2}{2} - \frac{1}{v} + C$

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

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

Substituting initial value gives C = 3/2

Is this correct? I don't have answers for practice exams.