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Q.
Let $y=f(x)$ satisfies the differential equation $\left(1+x^2\right) \frac{d y}{d x}+2 x y=2 x$ and $f(0)=2$. Then the number of integers in the range of $f(x)$ is
Differential Equations
Solution:
Given, $\frac{ dy }{ dx }+\left(\frac{2 x }{1+ x ^2}\right) y =\left(\frac{2 x }{1+ x ^2}\right)$ (Linear differential equation)
$\therefore $ I.F. $= e ^{\ln \left(1+ x ^2\right)}=1+ x ^2 . $
So, general solution is
$ y \cdot\left(1+x^2\right)=\int\left(\frac{2 x}{1+x^2}\right) \cdot\left(1+x^2\right) d x+C $
$\Rightarrow y\left(1+x^2\right)=x^2+C$
$\text { As } y(0)=2 \Rightarrow 2=0+c $
$\therefore y=f(x)=\left(\frac{x^2+2}{x^2+1}\right)=\left(1+\frac{1}{x^2+1}\right)$
Range of $f(x) = (1, 2]$.
