Question

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

Answer 1

(1+x2)$\frac {dy}{dx}$+2xy = $\frac {1}{1+x^2}$

⇒$\frac {dy}{dx}$+$\frac {2xy}{1+x^2}$ = $\frac {1}{(1+x^2)^2}$

This is a linear differential equation of the form:

$\frac {dy}{dx}$+py = Q (where p=$\frac {2x}{1+x^2}$ and Q=$\frac {1}{(1+x^2)^2}$

Now, I.F. = e∫pdx = $e^{∫\frac {2x}{1+x^2} dx}$= $e^{log(1+x^2)}$ = 1+x2
The general solution of the given differential equation is given by the relation,

y(I.F.) = ∫(Q×I.F.)dx + C

⇒y(1+x2) = ∫[$\frac {1}{(1+x^2)^2}$.(1+x2)]dx + C

⇒y(1+x2) = ∫$\frac {1}{(1+x^2)}$dx + C

⇒y(1+x2) = tan-1x+C ……...(1)

Now, y=0 at x=1.

Therefore,

0 = tan-1x+C

⇒C = -$\frac \pi4$

Substituting C=-$\frac \pi4$ in equation(1), we get:

y(1+x2) = tan-1x - $\frac \pi4$

This is the required general solution of the given differential equation.