# 2 $ \frac { d y } { d x } = \frac { x } { y } $ , $ y ( - 1 ) = 8 $

particular solution:
to advance in the circuit, find y when $ x = 1 $.

Question

# 2 $ \frac { d y } { d x } = \frac { x } { y } $ , $ y ( - 1 ) = 8 $

particular solution:
to advance in the circuit, find y when $ x = 1 $.

Accepted Answer

# 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 $\int t^n\,dt=\frac{t^{n + 1}}{n+1}+C$ ($n
eq - 1$), 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:
$y = 8$

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QUESTION IMAGE

Question

2 \\( \frac { d y } { d x } = \frac { x } { y } \\) , \\( y ( - 1 ) = 8 \\)

Explanation:

Step1: Separate variables

Step2: Integrate both sides

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

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

Answer: