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Find the general solution of the differential equation $x \log x\large\frac{dy}{dx}$$+y=\large\frac{2}{x}$$\log x$

1 Answer

  • To solve the first order linear differential equation of the form $\large\frac{dy}{dx}$$ + Py = Q$
  • (i) Write the given equation in the form of $\large\frac{dy}{dx}$$ + Py = Q$
  • (ii) Find the integrating factor (I.F) = $e^{\int Pdx}$.
  • (iii) Write the solution as y(I.F) = integration of $Q(I.F) dx + C$
Step 1:
Using the information in the tool box , let us write the equation as
$\large\frac{dy}{dx} +\frac{y}{x \log x} =\frac{ 2\log x}{x^2\log x} $ (by dividing throughout by $x logx).$
Here $P = \large\frac{1}{x\log x}$ and $Q = \large\frac{2\log x}{x^2}$
$\int p dx = \log(\log x)$, hence $e^{\int pdx} = e^{\large\log(\log x)} = \log x = I.F$
Step 2:
The solution is $y.\log x = \int \large\frac{2\log xdx}{x^2}$$ + C$
$y. \log x = 2 \int (\large\frac{1}{x^2}).$$\log x dx + C$
Integration of $(\large\frac{1}{x^2})$$\log x$ can be solved by integration by parts
Let $u = \log x$, then $du = (\large\frac{1}{x})$$dx$; $dv = (\large\frac{1}{x^2})$$dx$ hence $v = \large\frac{-1}{x}$
$2\int (\large\frac{1}{x^2})$$dx.\log x = 2[(\large\frac{-1}{x})\log x -\int (\large\frac{-1}{x})(\large\frac{1}{x})dx$
$\qquad\qquad\quad\quad\;\;= 2[(\large\frac{-1}{x})\log x +\int (\large\frac{1}{x^2})dx$
$\qquad\qquad\quad\quad\;\;= 2(\large\frac{-1}{x})$$\log x - (\large\frac{1}{x})$$ + C$
Hence the required solution is $y.\log x = 2[(\large\frac{-1}{x})$$ \log x - (\large\frac{1}{x}) $$+ C$
$y.\log x = \large\frac{-2}{x}$$[1 + \log x] + C$
This is the required solution.
answered Aug 1, 2013 by sreemathi.v