solve the equation.\n\frac{dy}{dx}=-\frac{2y^{2}}{x^{2}}\nchoose 1 answer:\na y=\frac{x}{-2 + cx}\nb…

Explanation:

Step1: Separate variables

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

Step2: Integrate both sides

Integrate $\int y^{-2}dy=-2\int x^{-2}dx$. Using the power - rule for integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$), we have $\frac{y^{-2 + 1}}{-2 + 1}=-2\frac{x^{-2+1}}{-2 + 1}+C$. Simplifying gives $-\frac{1}{y}=-2(-\frac{1}{x})+C$, which is $-\frac{1}{y}=\frac{2}{x}+C$.

Step3: Solve for y

Cross - multiply to get $-x = 2y+Cxy$. Then $-x=y(2 + Cx)$. So, $y=\frac{x}{-2 + Cx}$.

Answer:

A. $y=\frac{x}{-2 + Cx}$