Find a particular solution satisfying the given condition $(1+x^2)\large\frac{dy}{dx}$$+2xy=\large\frac{1}{1+x^2}$$;y=0\;when\;x=1$

$\begin{array}{1 1}(A)\;y(1+x^2) = \tan^{-1} x -\large\frac{ \pi}{2} \\(B)\;y(1+x^2) = \tan^{-1} x -\large\frac{ \pi}{4} \\(C)\;y(1+x^2) = \tan^{-1} x -\large\frac{ 2\pi}{3} \\ (D)\;y(1+x^2) = \tan^{-1} x +\large\frac{ \pi}{4} \end{array}$

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) = $\int\:Q(I.F) dx + C$
Step 1:
Using the information in the tool box, 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}$$. (1+x^2)\big] \:dx+ C \int\large\frac{ 1}{(1+x^2)}$$dx = \tan^{-1}x + C$
Hence the required solution is $y.(1+x^2) =\tan^{-1}x + C$
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.
edited Feb 11, 2014