Question

#1 $\frac{dy}{dx}=2x$, $y(1)=7$
particular solution:
to advance in the circuit, find y when x = 2.

Explanation:

Step1: Integrate the differential equation

Given \(\frac{dy}{dx}=2x\), integrate both sides with respect to \(x\). Using the power - rule \(\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C\) (\(n
eq - 1\)), we have \(y=\int 2x dx\).
Since \(\int 2x dx=2\times\frac{x^{2}}{2}+C=x^{2}+C\).

Step2: Use the initial condition to find \(C\)

We know that \(y(1) = 7\). Substitute \(x = 1\) and \(y=7\) into \(y=x^{2}+C\).
So \(7=1^{2}+C\), which gives \(C=7 - 1=6\).
The particular solution is \(y=x^{2}+6\).

Step3: Find \(y\) when \(x = 2\)

Substitute \(x = 2\) into \(y=x^{2}+6\).
\(y=2^{2}+6=4 + 6=10\).

Answer:

The particular solution is \(y=x^{2}+6\) and when \(x = 2\), \(y = 10\).