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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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Solve the following differential equation : $ (1+x^2)\large\frac{dy}{dx}$$+2xdy=\large\frac{1}{1+x^2},$ given y = 0 when x = 1.

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Toolbox:
  • To solve the first order linear differential equation of the form $\large\frac{dy}{dx}$$ + Py = Q$
  • Write the given equation in the form of $\large\frac{dy}{dx}$$ + Py = Q$
  • Find the integrating factor (I.F) = $e^{\int Pdx}$.
Step 1:
Let us rewrite the equation
dividing throughout by $(1+x^2)$ we get,
$\large\frac{dy}{dx } + \frac{2xy}{(1+x^2)}$$ = 1$
Here $P =\large\frac{ 2x}{(1+x^2)}$ and $Q = \large\frac{1}{(1+x^2)^2}$
Let us find the integrating factor(I.F)
$\int Pdx = \int\large\frac{ 2x}{(1+x^2)} $$= \log(1+x^2)$
Hence $I.F = e^{\log(1+x^2)} = (1+x^2)$
Step 2:
Hence the required solution is
$y.(1+x^2) =\int\big[\large\frac{1}{(1+x^2)^2}$$ . (1+x^2)\big] + C$
$\int\large\frac{ 1}{(1+x^2)}$$dx = \tan^{-1}x + C$
$y.(1+x^2) =\tan^{-1}x + C$
Step 3:
To evaluate the value of C
let us substitute the given values of $x = 1$ and $y =0$
$0 = \tan^{-1}(1) + C$
$\tan^{-1}(1)=\large\frac{\pi}{4}$
Therefore $C = \large\frac{- \pi}{4}$
substituting this we get,
$y(1+x^2) = \tan^{-1} x -\large\frac{ \pi}{4}$
This is the required solution.
answered Nov 8, 2013 by sreemathi.v
 

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