Question

4.c.) solve $\frac{dy}{dx}+\frac{1}{x}y = \frac{1}{x^{2}}$ $x>0$ and $y(1)=2$

Explanation:

Step1: Find the integrating factor

The given differential equation is of the form \(\frac{dy}{dx}+P(x)y = Q(x)\), where \(P(x)=\frac{1}{x}\) and \(Q(x)=\frac{1}{x^{2}}\).
The integrating factor \(I.F.=e^{\int P(x)dx}\).
Now, \(\int\frac{1}{x}dx=\ln x\), so \(I.F. = e^{\ln x}=x\).

Step2: Multiply the differential equation by the integrating factor

Multiply \(\frac{dy}{dx}+\frac{1}{x}y=\frac{1}{x^{2}}\) by \(x\).
We get \(x\frac{dy}{dx}+y=\frac{1}{x}\).
The left - hand side is the derivative of \(xy\) (by the product rule \((uv)^\prime = u^\prime v+uv^\prime\), here \(u = x\), \(v = y\), \((xy)^\prime=x\frac{dy}{dx}+y\)).
So, \(\frac{d(xy)}{dx}=\frac{1}{x}\).

Step3: Integrate both sides

Integrate \(\frac{d(xy)}{dx}=\frac{1}{x}\) with respect to \(x\).
\(\int d(xy)=\int\frac{1}{x}dx\).
\(xy=\ln x + C\).

Step4: Use the initial condition

Given \(y(1) = 2\). Substitute \(x = 1\) and \(y = 2\) into \(xy=\ln x + C\).
\(1\times2=\ln(1)+C\). Since \(\ln(1)=0\), then \(C = 2\).

Step5: Solve for \(y\)

From \(xy=\ln x + 2\), we can solve for \(y\).
\(y=\frac{\ln x + 2}{x}\).

Answer:

\(y=\frac{\ln x + 2}{x}\)