Question

4a.) solve $y+\frac{1}{x}y = 2$

Explanation:

Step1: Identify the type of differential equation

This is a first - order linear differential equation of the form \(y'+P(x)y = Q(x)\), where \(P(x)=\frac{1}{x}\) and \(Q(x) = 2\).

Step2: Find the integrating factor

The integrating factor \(\mu(x)=e^{\int P(x)dx}\).
Since \(\int\frac{1}{x}dx=\ln|x|\) (for \(x
eq0\)), then \(\mu(x)=e^{\ln|x|}=|x|\). Assuming \(x>0\) (a similar analysis can be done for \(x < 0\)), \(\mu(x)=x\).

Step3: Multiply the differential equation by the integrating factor

Multiply \(y'+\frac{1}{x}y = 2\) by \(x\):
\(x y'+y=2x\).
Notice that the left - hand side is the derivative of the product \(xy\) by the product rule \((uv)'=u'v + uv'\) (here \(u = x\), \(v=y\), \(u'=1\), \(v'=y'\)). So \((xy)'=2x\).

Step4: Integrate both sides

Integrate \((xy)'=2x\) with respect to \(x\).
\(\int(xy)'dx=\int 2x dx\).
Using the fundamental theorem of calculus \(\int(xy)'dx=xy\) and \(\int 2x dx=x^{2}+C\) (where \(C\) is the constant of integration).
So \(xy=x^{2}+C\).

Step5: Solve for \(y\)

Divide both sides of \(xy=x^{2}+C\) by \(x\) (for \(x
eq0\)):
\(y=x+\frac{C}{x}\).

Answer:

\(y=x+\frac{C}{x}\)