Question

Solve (1+y2)dx = (tan-1y - x)dy

Answer 1

hope this helps.......

Answer 2

Answer:

By solving we get, $\left(\tan ^{-1} y-1\right) e^{\tan ^-1} y+c$

Step-by-step explanation:

The given differential equation is

$&\left(1+y^{2}\right) d x=\left(\tan ^{-1} y-x\right) d y$

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

which is a linear differential equation.

$\therefore \mathrm{IF}=e^{\int \frac{d y}{1+y^{2}}}=e^{\tan -1} y$

Hence, the solution is

$&x \cdot e^{\tan -1} y=\int \frac{\tan ^{-1} y e^{\tan -1} y}{1+y^{2}} d y+c$

$&=\int t e^{t} d t+c, t=\tan ^{-1} y$

$&=(t-1) e^{t}+c$

$&=\left(\tan ^{-1} y-1\right) e^{\tan ^-1} y+c$

which is the required solution.