Question

Solve the differential equation $({\tan }^{-1}x-y)dx=(1+x^2)dy$.

Answer 1

$(ta{n}^{-1}x-y)dx=(1+x^2)dy$

$\therefore \frac{dy}{dx}=\frac{{\tan }^{-1}x-y}{1+x^2}$

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

Above equation is linear $\frac{dy}{dx}+Py=Q$

$P=\frac{1}{1+x^2}Q=\frac{{\tan }^{-1}x}{1+x^2}$

IF = e^{\int Pdx}$

$=e^{\int (1/1+x^2)dx}$

$=e^{{\tan }^{-1}x}$

solution is given as

$y{e}^{-{\tan }^{-1}x}=\int \left(\frac{{\tan }^{1}x}{1+x^2}{e}^{{\tan }^{-1}x}\right)dx+c$ ...(1)

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

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

$\int t{e}^{t}dt$

$=t\times \int e^{t}dt-\int \left(\frac{d}{dt}\right(t)\times \int e^{t}dt]at$

$=t{e}^t-\int e^{t}dt$

$=t{e}^t-e^t$

$={\tan }^{-1}x{e}^{{\tan }^{-1}x}-e^{{\tan }^{-1}x}$

From (1), we get

$y{e}^{{\tan }^{-1}x}={\tan }^{-1}x{e}^{{\tan }^{-1}x}-e^{{\tan }^{-1}x}+C=y={\tan }^{-1}x-1+c{e}^{{\tan }^{-1}x}$