Question

The given differential equation is: $(x+3y^2)\frac{dy}{dx} = y$

Answer 1

The general solution to the differential equation is: $x = 3y^2 + Cy$

Answer 2

The given differential equation is $(x+3y^2)\frac{dy}{dx} = y$. To solve this, we can rewrite it as $\frac{dx}{dy} = \frac{x+3y^2}{y}$, which simplifies to $\frac{dx}{dy} = \frac{x}{y} + 3y$. Rearranging this into the standard form of a first-order linear differential equation, $\frac{dx}{dy} + P(y)x = Q(y)$, we get: $\frac{dx}{dy} - \frac{1}{y}x = 3y$ Here, $P(y) = -\frac{1}{y}$ and $Q(y) = 3y$.

The integrating factor (IF) is calculated as $IF = e^{\int P(y) dy}$. $IF = e^{\int (-\frac{1}{y}) dy} = e^{-\ln|y|} = e^{\ln|y|^{-1}} = |y|^{-1}$. Assuming $y \neq 0$, we take $IF = \frac{1}{y}$.

The general solution is given by $x \cdot IF = \int Q(y) \cdot IF \, dy + C$. Substituting the values: $x \cdot \frac{1}{y} = \int (3y) \cdot \left(\frac{1}{y}\right) dy + C$ $\frac{x}{y} = \int 3 \, dy + C$ $\frac{x}{y} = 3y + C$ $x = y(3y + C)$ $x = 3y^2 + Cy$