# 4 $y=-4y$ and $y(0)=8$.\nparticular solution: \nto advance in the circuit, find $y(2)$.
# 4 $y=-4y$ and $y(0)=8$.\nparticular solution: \nto advance in the circuit, find $y(2)$.
Answer
Explanation:
Step1: Solve the differential equation
Separate variables: $\frac{dy}{y}=-4dx$. Integrate both sides: $\int\frac{dy}{y}=\int - 4dx$. Using $\int\frac{1}{u}du=\ln|u|+C$, we get $\ln|y|=-4x + C$. Exponentiate both sides: $y = e^{-4x + C}=e^{C}e^{-4x}$. Let $k = e^{C}$, so $y = ke^{-4x}$.
Step2: Use the initial condition
Given $y(0)=8$, substitute $x = 0$ and $y = 8$ into $y = ke^{-4x}$. We have $8=k\cdot e^{0}$, so $k = 8$. The particular solution is $y = 8e^{-4x}$.
Step3: Find $y(2)$
Substitute $x = 2$ into $y = 8e^{-4x}$. $y(2)=8e^{-4\times2}=8e^{-8}$.
Answer:
The particular solution is $y = 8e^{-4x}$ and $y(2)=8e^{-8}$.