Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  CBSE XII  >>  Math  >>  Differential Equations
0 votes

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

Can you answer this question?

1 Answer

0 votes
  • 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

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App