#______ xy\\(\\frac{dy}{dx}=lnx\\) y(1)= -2\n\nparticular solution:__________________________\nto advance in…
#______ xy\\(\\frac{dy}{dx}=lnx\\) y(1)= -2\n\nparticular solution:__________________________\nto advance in the circuit, find x when y= -\\(\\sqrt{5}\\).
Answer
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}).