Question

#______ xy\\(\frac{dy}{dx}=lnx\\) y(1)= -2

particular solution:__________________________
to advance in the circuit, find x when y= -\\(\sqrt{5}\\).

Explanation:

Step1: Separate variables

Given \(xy\frac{dy}{dx}=\ln x\), we can rewrite it as \(y\;dy=\frac{\ln x}{x}\;dx\).

Step2: Integrate both sides

Integrate \(\int y\;dy=\int\frac{\ln x}{x}\;dx\).
For \(\int y\;dy\), using the power - rule \(\int y\;dy=\frac{y^{2}}{2}+C_1\).
Let \(u = \ln x\), then \(du=\frac{1}{x}dx\). So \(\int\frac{\ln x}{x}\;dx=\int u\;du=\frac{u^{2}}{2}+C_2=\frac{(\ln x)^{2}}{2}+C_2\).
The general solution is \(\frac{y^{2}}{2}=\frac{(\ln x)^{2}}{2}+C\).

Step3: Use the initial condition \(y(1)= - 2\)

Substitute \(x = 1\) and \(y=-2\) into \(\frac{y^{2}}{2}=\frac{(\ln x)^{2}}{2}+C\).
\(\frac{(-2)^{2}}{2}=\frac{(\ln1)^{2}}{2}+C\). Since \(\ln1 = 0\), we have \(C = 2\).
The particular solution is \(\frac{y^{2}}{2}=\frac{(\ln x)^{2}}{2}+2\), or \(y^{2}=(\ln x)^{2}+4\).

Step4: Find \(x\) when \(y =-\sqrt{5}\)

Substitute \(y =-\sqrt{5}\) into \(y^{2}=(\ln x)^{2}+4\).
\((-\sqrt{5})^{2}=(\ln x)^{2}+4\), so \(5=(\ln x)^{2}+4\).
Then \((\ln x)^{2}=1\), which gives \(\ln x=\pm1\).
If \(\ln x = 1\), then \(x = e\); if \(\ln x=-1\), then \(x = e^{-1}=\frac{1}{e}\).

Answer:

The particular solution is \(y^{2}=(\ln x)^{2}+4\). When \(y =-\sqrt{5}\), \(x = e\) or \(x=\frac{1}{e}\).