Question

1. the function ( y = f(x) ) satisfies the differential equation ( \frac{dy}{dx} = 2xy ) with the initial condition ( f(0) = 5 ). what is the value of ( f(2)? )

Explanation:

Step1: Separate variables

Given \(\frac{dy}{dx}=2xy\), we can rewrite it as \(\frac{dy}{y}=2x dx\) (separate \(y\) and \(x\) terms).

Step2: Integrate both sides

Integrate \(\int\frac{dy}{y}=\int 2x dx\).
Using the integral formulas \(\int\frac{1}{u}du=\ln|u| + C\) (\(u = y\)) and \(\int ax^{n}dx=\frac{ax^{n + 1}}{n+1}+C\) (\(a = 2\), \(n=1\)), we get \(\ln|y|=x^{2}+C\).

Step3: Solve for \(y\)

Exponentiating both sides, \(y = e^{x^{2}+C}=e^{C}e^{x^{2}}\). Let \(k = e^{C}\), so \(y=ke^{x^{2}}\).

Step4: Use the initial condition

Given \(f(0)=5\), substitute \(x = 0\) and \(y = 5\) into \(y=ke^{x^{2}}\).
We have \(5=ke^{0}\), since \(e^{0}=1\), then \(k = 5\). So the function is \(y = 5e^{x^{2}}\).

Step5: Find \(f(2)\)

Substitute \(x = 2\) into \(y = 5e^{x^{2}}\). Then \(y=5e^{2^{2}}=5e^{4}\).

Answer:

\(5e^{4}\)