17. if $y(0)= - 2$, what is the particular solution for $\frac{dy}{dx}=\frac{6x^{2}}{y}$?\n a. $y =…

Explanation:

Step1: Separate variables

We start with $\frac{dy}{dx}=\frac{6x^{2}}{y}$. Cross - multiply to get $y;dy = 6x^{2};dx$.

Step2: Integrate both sides

Integrate $\int y;dy=\int6x^{2};dx$. Using the power rule $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$), we have $\frac{y^{2}}{2}=6\times\frac{x^{3}}{3}+C$, which simplifies to $\frac{y^{2}}{2}=2x^{3}+C$.

Step3: Use the initial condition

Given $y(0)=-2$. Substitute $x = 0$ and $y=-2$ into $\frac{y^{2}}{2}=2x^{3}+C$. Then $\frac{(-2)^{2}}{2}=2\times0^{3}+C$, so $C = 2$.

Step4: Solve for y

We have $\frac{y^{2}}{2}=2x^{3}+2$, then $y^{2}=4x^{3}+4$. Since $y(0)=-2$, $y=-\sqrt{4x^{3}+4}$.

Answer:

d. $y =-\sqrt{4x^{3}+2}$ (There is a small error in the original problem - the correct general form after solving should be $y =-\sqrt{4x^{3}+4}$, but among the given options, the closest one in form is d. with a possible mis - typing in the constant term in the options)