The teacher hasn't uploaded a solution for this question yet.
The given differential equation is $y~dx = (x + 2y^2)~dy$. We can rewrite this as: $$\frac{dx}{dy} = \frac{x + 2y^2}{y} = \frac{x}{y} + 2y$$ $$\frac{dx}{dy} - \frac{1}{y}x = 2y$$
This is a linear differential equation of the form $\frac{dx}{dy} + P(y)x = Q(y)$, where $P(y) = -\frac{1}{y}$ and $Q(y) = 2y$.
The integrating factor is given by $IF = e^{\int P(y)~dy}$. $$IF = e^{\int -\frac{1}{y}~dy} = e^{-\ln|y|} = e^{\ln|y^{-1}|} = \frac{1}{y}$$
The general solution is given by: $$x \cdot IF = \int Q(y) \cdot IF ~dy + C$$ $$x \cdot \frac{1}{y} = \int 2y \cdot \frac{1}{y} ~dy + C$$ $$\frac{x}{y} = \int 2 ~dy + C$$ $$\frac{x}{y} = 2y + C$$ $$x = 2y^2 + Cy$$
The general solution of the given differential equation is $x = 2y^2 + Cy$.
Final Answer: $x = 2y^2 + Cy$
AI generated content. Review strictly for academic accuracy.