dy/dx = -(x + 6)/7y^2\nchoose 1 answer:\na y = ±√(-(x + 6)/7 + c)\nb y = c√(-(x + 6)/7)\nc y = c∛(-3x^2/14…

Explanation:

Step1: Separate variables

Separate the variables in the differential - equation $\frac{dy}{dx}=-\frac{x + 6}{7y^{2}}$. We get $7y^{2}dy=-(x + 6)dx$.

Step2: Integrate both sides

Integrate $\int7y^{2}dy=-\int(x + 6)dx$. For the left - hand side, using the power rule $\int y^{n}dy=\frac{y^{n + 1}}{n+1}+C$ ($n\neq - 1$), we have $\int7y^{2}dy=7\times\frac{y^{3}}{3}=\frac{7y^{3}}{3}$. For the right - hand side, $\int(x + 6)dx=\int xdx+\int6dx=\frac{x^{2}}{2}+6x+C_1$. So, $-\int(x + 6)dx=-\frac{x^{2}}{2}-6x + C$. Then $\frac{7y^{3}}{3}=-\frac{x^{2}}{2}-6x + C$.

Step3: Solve for y

Multiply both sides by $\frac{3}{7}$: $y^{3}=\frac{3}{7}\left(-\frac{x^{2}}{2}-6x + C\right)=-\frac{3x^{2}}{14}-\frac{18x}{7}+C'$ (where $C'=\frac{3C}{7}$). Take the cube - root of both sides: $y=\sqrt[3]{-\frac{3x^{2}}{14}-\frac{18x}{7}+C'}$.

Answer:

D. $y=\sqrt[3]{-\frac{3x^{2}}{14}-\frac{18x}{7}+C}$