#1 $\\frac{dy}{dx}=2x$, $y(1)=7$\nparticular solution: \nto advance in the circuit, find y when x = 2.
#1 $\\frac{dy}{dx}=2x$, $y(1)=7$\nparticular solution: \nto advance in the circuit, find y when x = 2.
Answer
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\neq - 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).