Question

\(\;{\bf{9}} - {\bf{14}}\) Find the solution of the differential equation that satisfies the given initial condition.

\(\frac{{dy}}{{dx}} = \frac{{\ln x}}{{xy}},\;\;\;y(1) = 2\)

Short answer

\(y = \sqrt {{{(\ln x)}^2} + 4} {\rm{. }}\)

Step-by-step solution

Definition

A differential equation is an equation with one or more derivatives of a function.

Separate variables

Consider the differential equation\(\frac{{dy}}{{dx}} = \frac{{\ln x}}{{xy}}\)with initial condition\(y(1) = 2\).

By using variable separable method,

\(\begin{aligned}\frac{{dy}}{{dx}} &= \frac{{\ln x}}{{xy}}\\xydy &= \ln xdx\\ydy &= \frac{{\ln x}}{x}dx\end{aligned}\)

Integrate

Integrating both sides we have:

\(\int y dy = \int {\frac{{\ln x}}{x}} dx + C\)

Using integral by substitution, let\(u = \ln x\), and\(du = \frac{1}{x}dx\).

\(\int y dy = \int u du + C\)

\(\frac{{{y^{1 + 1}}}}{{1 + 1}} = \frac{{{u^{1 + 1}}}}{{1 + 1}} + C\;\) Using formula\(\int {{x^n}} dx = \frac{{{x^{n + 1}}}}{{n + 1}} + C,n \ne - 1\)

\(\frac{1}{2}{y^2} = \frac{1}{2}{u^2} + C\)

\(\frac{1}{2}{y^2} = \frac{1}{2}{(\ln x)^2} + C\;\;\;\)Since\(u = \ln x\) ------(1)

Apply initial condition

Using initial condition\(y(1) = 2\) that means\(x = 1,y = 2\).

\(\begin{aligned}\frac{1}{2} \cdot {2^2} &= \frac{1}{2}{(\ln 1)^2} + C\\\frac{4}{2} &= \frac{1}{2}{(0)^2} + C\\C &= 2\end{aligned}\)

Find solution

Substitute value\(C = 2\) in equation (1) we have:

\(\begin{aligned}\frac{1}{2}{y^2} &= \frac{1}{2}{(\ln x)^2} + 2\;\;\;{\rm{ Since }}C = 2\\\frac{1}{2}{y^2} &= \frac{1}{2}\left( {{{(\ln x)}^2} + 4} \right)\\{y^2} &= {(\ln x)^2} + 4\\y &= \pm \sqrt {{{(\ln x)}^2} + 4} \end{aligned}\)

Since, \(y(1) = 2 > 0\)we have\(y = \sqrt {{{(\ln x)}^2} + 4} \)

Therefore, the solution of given the differential equations is\(y = \sqrt {{{(\ln x)}^2} + 4} {\rm{. }}\)