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Home  >>  CBSE XII  >>  Math  >>  Differential Equations
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Find the general solution of the differential equation $(1+x^2)dy+2xy\;dx=\cot x\;dx\;(x\neq0)$

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Toolbox:
  • 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:
By using the information in the tool box, let us write the equation as
$\large\frac{dy}{dx} +\frac{2xy}{(1+x^2) }=\frac{ \cot x}{(1+x^2)}$ ( divide throughout by $(1+x^2)dx $
Here $P = \large\frac{2x}{(1+x^2)}$ and $Q =\large\frac{ \cot x}{(1+x^2)}$
$\int pdx = \int\large\frac{ 2x}{(1+x^2) }=$$ \log(1+x^2)$
Hence the $I.F = e^{\large\log(1+x^2) }= (1+x^2)$
Hence the solution is $y.(1+x^2) = \int [\cot x.(1+x^2).(1+x^2)dx] + C$
Step 2:
$ \int \cot xdx = \log|\sin x| + C$
$y.(1+x^2) = \log|\sin x| + C$
This is the required solution.
answered Aug 1, 2013 by sreemathi.v
 

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