Question

For the differential equation given in the question find a particular solution satisfying the given condition.

$(1+x^2)\frac{dy}{dx}+2xy=\frac{1}{1+x^2},$where y=0 and x=1.

Answer 1

Given, $(1+x^2)\frac{dy}{dx}+2xy=\frac{1}{1+x^2}$
On dividing by $(1+x^2)$ both sides, we get
$\frac{dy}{dx}+\frac{2xy}{1+x^2}=\frac{1}{(1+x^2{)}^2}$
On comparing with the form $\frac{dy}{dx}+Py=Q,$ we get
$P=\frac{2x}{1+x^2},Q=\frac{1}{(1+x^2{)}^2}$
$\therefore IF=e^{\int \frac{2x}{1+x^{2}dx}}$
Put $1+x^2=t\Rightarrow 2x=\frac{dt}{dx}\Rightarrow dx=\frac{dt}{2x}$
$\therefore IF=e^{\int \frac{2x}{t}\times \frac{dt}{2x}}\Rightarrow IF=e^{log|t|}=t\Rightarrow IF=1+x^2$
The general solution of the given equation is given by
$y.IF=\int Q\times IFdx+C\Rightarrow y(1+x^2)=\int (\frac{1}{(1+x^2{)}^2}\times (1+x^2\left)\right)dx+C\Rightarrow y(1+x^2)=\int \frac{1}{1+x^2}dx+C\Rightarrow y(1+x^2)=ta{n}^{-1}x+C$
Now, y=0 and x=1, then
$0(1+1)=ta{n}^{-1}\left(1\right)+C\Rightarrow 0=\frac{\pi }{4}+C\Rightarrow C=-\frac{\pi }{4}$
On putting the value of C in Eq. (ii), we get
$(1+x^2)y=ta{n}^{-1}x-\frac{\pi }{4}$
Which is the required solution of the given differential equation.