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Question

(C) Given the differential equation: $(y^3 + x) \frac{dy}{dx} = y$. Rearranging the equation: $\frac{dx}{dy} = \frac{y^3 + x}{y} = y^2 + \frac{x}{y}$.

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$8$

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(C) Given the differential equation: $(y^3 + x) \frac{dy}{dx} = y$. Rearranging the equation: $\frac{dx}{dy} = \frac{y^3 + x}{y} = y^2 + \frac{x}{y}$. This is a linear differential equation in $x$: $\frac{dx}{dy} - \frac{1}{y}x = y^2$. The integrating factor is $IF = e^{\int -\frac{1}{y} dy} = e^{-\ln|y|} = \frac{1}{y}$. Multiplying by $IF$: $\frac{1}{y} \frac{dx}{dy} - \frac{1}{y^2} x = y$. Integrating both sides with respect to $y$: $\int \frac{d}{dy} (\frac{x}{y}) dy = \int y dy$. $\frac{x}{y} = \frac{y^2}{2} + C$. $x = \frac{y^3}{2} + Cy \implies 2x = y^3 + 2Cy$. Since $y^3 = ax + b$,we rewrite as $y^3 = 2x - 2Cy$. Comparing $y^3 = ax + b$,we see $a = 2$ and $b = -2Cy$. Using $y(4) = 2$: $2^3 = 2(4) + b \implies 8 = 8 + b \implies b = 0$. Then $y^3 = 2x + 0$,so $a = 2$ and $b = 0$. Calculating $4a + 12b^2 = 4(2) + 12(0)^2 = 8 + 0 = 8$.

If the form of the solution of the differential equation $(y^3+x) \frac{dy}{dx} = y$ with the condition $y(4) = 2$ is $y^3 = ax + b$,then $4a + 12b^2 = $

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