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Home  >>  CBSE XII  >>  Math  >>  Differential Equations
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Find the general solution of the differential equation $x\large\frac{dy}{dx}$$+y-x+xy\cot x=0\;(x\neq0)$

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Toolbox:
  • The first order linear equation can be solved as:
  • (i) By writing the equation as $\large\frac{dy}{dx}$$ +Py = Q$
  • (ii) Find the integral factor(I.F) $\int Pdx = e^{\int Pdx}$
  • (iii) Write the solution as $y. (I.F) = \int Q(I.F)dx + C$
Step 1:
Using the information in the tool box, let us write the equation as $\large\frac{dy}{dx} =\frac{ (1+x\cot x)y}{x} $$= 1 $
Here $P = (\large\frac{1}{x})$$ + \cot x$ and $Q = 1$
$\int pdx = \int [\large\frac{1}{x}$$ + \cot x]dx$
$\qquad\;\;= \log x + \log|\sin x|$
$I.F = e^[\log\sin x + \log x] = e^{\log[x\sin x] }= [x\sin x]$
The required solution is $y.\log[x\sin x] = \int1.[x\sin x] + C$
Step 2:
Integration of $[x\sin x]$ can be done by parts.
Let $u = x$ then $du = dx$ and $dv = \sin xdx, v = -\cos x$
$\int x\sin x = x(-\cos x) - \int(-\cos x).dx$
$\qquad\;\;\;\;\;= -[x\cos x] +\int\cos xdx$
$\qquad\;\;\;\;\;= -[x\cos x] +\sin x + C$
Dividing throughout by $x\sin x$ we get
$y = -\cot x + \large\frac{1}{x} $$+ C$
This is the required solution.
answered Jul 31, 2013 by sreemathi.v
 

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