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.