Question

4 $y=-4y$ and $y(0)=8$.

particular solution:
to advance in the circuit, find $y(2)$.

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 $y(2)=8e^{-8}$.