Question

Find the particular solution of the differential equation $(1+x^2)\frac{dy}{dx}=(e^{m{\tan }^{-1}x}-y)$, given that $y=1$, when $x=0$.

Answer 1

$(1+x^2)\frac{dy}{dx}=(e^{m{\tan }^{-1x}}-y)$

$\Rightarrow \frac{dy}{dx}+\frac{1}{1+x^2}y=\frac{e^{m{\tan }^{-1}}x}{1+x^2}...\left(i\right)$

Here, $P=\frac{1}{1+x^2}$ and $Q=\frac{e^{m{\tan }^{-1}}x}{1+x^2}$

I.F. = $e^{\int \frac{1}{1+x^2}dx}=e^{{\tan }^{-1}x}$

$y.e^{{\tan }^{-1}x}=\int e^{{\tan }^{-1}x}.\frac{e^{m{\tan }^{-1}x}}{1+x^2}dx+c$

= $I+c$

Put ${\tan }^{-1}x=t$

$\Rightarrow \frac{1}{1+x^2}dx=dt$

$I=\int e^t.e^{mt}dt$

= $\int e^{(m+1)t}dt$

= $\frac{e^{(m+1)t}}{m+1}=\frac{e^{(m+1){\tan }^{-1}x}}{m+1}$

$\therefore y.e^{{\tan }^{-1}x}=\frac{e^{(m+1){\tan }^{-1}x}}{m+1}+c...\left(ii\right)$

When $x=0,y=1$, we have

$1.e^0=\frac{e^0}{m+1}+c$

$\Rightarrow c=\frac{m}{m+1}$

Put the value of c in eq. (ii)
$\therefore y.e^{{\tan }^{-1}x}=\frac{e^{(m+1){\tan }^{-1}x}}{m+1}+\frac{m}{m+1}$

$\Rightarrow y.(m+1)e^{{\tan }^{-1}x}=e^{(m+1){\tan }^{-1}x}+m$