# 2 \\( \\frac { d y } { d x } = \\frac { x } { y } \\) , \\( y ( - 1 ) = 8 \\)\n\nparticular solution: \nto…

Explanation:

Step1: Separate variables

We have (\frac{dy}{dx}=\frac{x}{y}), which can be rewritten as (y,dy = x,dx).

Step2: Integrate both sides

Integrating (\int y,dy=\int x,dx). Using the power - rule (\int t^n,dt=\frac{t^{n + 1}}{n+1}+C) ((n\neq - 1)), we get (\frac{y^{2}}{2}=\frac{x^{2}}{2}+C).

Step3: Use the initial condition (y(-1)=8)

Substitute (x=-1) and (y = 8) into (\frac{y^{2}}{2}=\frac{x^{2}}{2}+C). [ \begin{align*} \frac{8^{2}}{2}&=\frac{(-1)^{2}}{2}+C\ \frac{64}{2}&=\frac{1}{2}+C\ 32&=\frac{1}{2}+C\ C&=32-\frac{1}{2}=\frac{64 - 1}{2}=\frac{63}{2} \end{align*} ] So the equation is (\frac{y^{2}}{2}=\frac{x^{2}}{2}+\frac{63}{2}), or (y^{2}=x^{2}+63).

Step4: Find (y) when (x = 1)

Substitute (x = 1) into (y^{2}=x^{2}+63). Then (y^{2}=1 + 63=64), so (y=\pm8). Since (y(-1)=8) (and the function is continuous in the domain of the differential equation), we take (y = 8).

Answer:

(y = 8)