Solve the equation. (dy)/(dx)=-(x+6)/(7y^(2)) Choose 1 answer: (A) y=+-sqrt(-(x+6)/(7)+C) (B) y=Csqrt(-(x+6)/(7)) (C) y=Croot(3)(-(3x^(2))/(14)-(18 x)/(7)) (D) y=root(3)(-(3x^(2))/(14)-(18 x)/(7)+C)

Separate variables: We are given the differential equation (dy)/(dx) = -(x+6)/(7y^2). This is a separable differential equation, which means we can separate the variables y and x on different sides of the equation. To do this, we multiply both sides by 7y^2 dy and dx to get 7y^2 dy = -(x+6) dx.Integrate both sides: Next, we integrate both sides of the equation. The left side with respect to y, and the right side with respect to x. This gives us the integral of 7y^2 dy = integral of -(x+6) dx.Combine constants: The integral of 7y^2 dy is (7/3)y^3 + C1, where C1 is the constant of integration. The integral of -(x+6) dx is -(1/2)x^2 - 6x + C2, where C2 is another constant of integration.Solve for y: Since the constants of integration on both sides of the equation are arbitrary, we can combine them into a single constant C. Therefore, we have (7/3)y^3 = -(1/2)x^2 - 6x + C.Final solution: To solve for y, we divide both sides by (7/3) to get y^3 = -((1/2)x^2 + 6x)/(7/3) + C/(7/3). Simplifying the right side, we get y^3 = -(3/14)x^2 - (18/7)x + C/(7/3).Taking the cube root of both sides to solve for y, we get y = cube root of (-(3/14)x^2 - (18/7)x + C/(7/3)). This matches answer choice (D) if we consider the constant C/(7/3) as a new constant C'.

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Find derivatives of using multiple formulae

Full solution

Q. Solve the equation.

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\frac{d y}{d x}=-\frac{x+6}{7 y^{2}}

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Choose

1

answer:

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(A)

y= \pm \sqrt{-\frac{x+6}{7}+C}

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(B)

y=C \sqrt{-\frac{x+6}{7}}

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(C)

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

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(D)

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

Separate variables: We are given the differential equation $(\frac{dy}{dx}) = -\frac{(x+6)}{7y^2}$ . This is a separable differential equation, which means we can separate the variables $y$ and $x$ on different sides of the equation. To do this, we multiply both sides by $7y^2 dy$ and $dx$ to get $7y^2 dy = -(x+6) dx$ .
Integrate both sides: Next, we integrate both sides of the equation. The left side with respect to $y$ , and the right side with respect to $x$ . This gives us the integral of $7y^2 \, dy =$ integral of $- (x+6) \, dx$ .
Combine constants: The integral of $7y^2 \, dy$ is $(\frac{7}{3})y^3 + C_1$ , where $C_1$ is the constant of integration. The integral of $-(x+6) \, dx$ is $-(\frac{1}{2})x^2 - 6x + C_2$ , where $C_2$ is another constant of integration.
Solve for $y$ : Since the constants of integration on both sides of the equation are arbitrary, we can combine them into a single constant $C$ . Therefore, we have $\frac{7}{3}y^3 = -\frac{1}{2}x^2 - 6x + C$ .
Final solution: To solve for $y$ , we divide both sides by $(7/3)$ to get $y^3 = -\left(\frac{1}{2}x^2 + 6x\right)/(7/3) + \frac{C}{(7/3)}$ . Simplifying the right side, we get $y^3 = -\left(\frac{3}{14}\right)x^2 - \left(\frac{18}{7}\right)x + \frac{C}{(7/3)}$ .
Final solution: To solve for $y$ , we divide both sides by $(7/3)$ to get $y^3 = -\left(\frac{1}{2}x^2 + 6x\right)/(7/3) + \frac{C}{(7/3)}$ . Simplifying the right side, we get $y^3 = -\left(\frac{3}{14}\right)x^2 - \left(\frac{18}{7}\right)x + \frac{C}{(7/3)}$ .Taking the cube root of both sides to solve for $y$ , we get $y = \text{cube root of } \left(-\left(\frac{3}{14}\right)x^2 - \left(\frac{18}{7}\right)x + \frac{C}{(7/3)}\right)$ . This matches answer choice (D) if we consider the constant $\frac{C}{(7/3)}$ as a new constant $C'$ .