2. given the differential equation dy/dx=-2x/y\n\n i. on the axes provided, sketch a slope - field for the…

Explanation:

Step1: Recall slope - field concept

The slope of the tangent line at a point $(x,y)$ for the differential equation $\frac{dy}{dx}=-\frac{2x}{y}$ is given by $m = -\frac{2x}{y}$. For each of the twelve points in the table, substitute the $x$ and $y$ values into the formula for $\frac{dy}{dx}$ to find the slope. For example, when $x = 0,y=1$, $\frac{dy}{dx}=-\frac{2\times0}{1}=0$.

Step2: Find the slope of the tangent line for part (ii)

We know that the slope of the tangent line to the curve $y = f(x)$ at the point $(x_0,y_0)$ is given by $m=\frac{dy}{dx}\big|{(x_0,y_0)}$. Given the differential equation $\frac{dy}{dx}=-\frac{2x}{y}$ and the point $(1, - 1)$, substitute $x = 1$ and $y=-1$ into the differential equation: $\frac{dy}{dx}\big|{(1,-1)}=-\frac{2\times1}{-1}=2$. The equation of the tangent - line using the point - slope form $y - y_0=m(x - x_0)$ with $(x_0,y_0)=(1,-1)$ and $m = 2$ is $y+1=2(x - 1)$, or $y=2x-3$. To approximate $f(1.1)$, substitute $x = 1.1$ into the tangent - line equation: $y=2\times1.1-3=2.2 - 3=-0.8$.

Step3: Solve the differential equation for part (iii)

We have the separable differential equation $\frac{dy}{dx}=-\frac{2x}{y}$. Rearrange it as $y;dy=-2x;dx$. Integrate both sides: $\int y;dy=-\int 2x;dx$. Using the power rule for integration $\int x^n;dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$), we get $\frac{y^{2}}{2}=-x^{2}+C$. Given the initial condition $f(1)=-1$, substitute $x = 1$ and $y=-1$ into the equation $\frac{y^{2}}{2}=-x^{2}+C$: $\frac{(-1)^{2}}{2}=-1^{2}+C$. Simplify: $\frac{1}{2}=-1 + C$, so $C=\frac{3}{2}$. The particular solution is $\frac{y^{2}}{2}=-x^{2}+\frac{3}{2}$, or $y^{2}=-2x^{2}+3$.

Answer:

i. Sketch the slope - field by calculating the slopes at the given points using $\frac{dy}{dx}=-\frac{2x}{y}$. ii. The equation of the tangent line is $y = 2x-3$ and $f(1.1)\approx - 0.8$. iii. The particular solution is $y^{2}=-2x^{2}+3$.