Question

1. $\frac{dy}{dx}=y + 2$, $y(0)=2$

Explanation:

Step1: Separate variables

Separate the variables in the differential equation \(\frac{dy}{dx}=y + 2\). We get \(\frac{dy}{y + 2}=dx\).

Step2: Integrate both sides

Integrate both sides of the equation \(\int\frac{dy}{y + 2}=\int dx\).
Using the formula \(\int\frac{1}{u}du=\ln|u|+C\) (where \(u = y + 2\) for the left - hand side and for the right - hand side \(\int dx=x + C\)), we have \(\ln|y + 2|=x + C\).

Step3: Solve for \(y\)

Exponentiate both sides to get rid of the natural logarithm: \(y+2 = e^{x + C}\).
Since \(e^{x + C}=e^{C}e^{x}\), let \(k = e^{C}\), then \(y=ke^{x}-2\).

Step4: Use the initial condition

Use the initial condition \(y(0) = 2\). Substitute \(x = 0\) and \(y = 2\) into \(y=ke^{x}-2\).
We get \(2=k e^{0}-2\), since \(e^{0}=1\), then \(2=k - 2\), so \(k = 4\).

Answer:

\(y = 4e^{x}-2\)