Solve (1 + y^2) dx + (x - e ^-tan^-1y) dy = 0, given that when y =o, then x = 0.

Question

Solve (1 + y²) dx + (x - \(e^{-tan^{-1}}\)_^y) dy = 0, given that when y = 0, then x = 0.

Answer 1

Given Differential Equation :

(1 + y²) dx + (x - e ^-tan-1_y) dy = 0

Formula :

i) \(\int \frac{1}{(1+x^2)}\) dx = tan^-1x

ii) General solution :

For the differential equation in the form of

\(\frac{dx}{dy} \, + Py \, =Q\)

General solution is given by,

x. (I.F.) = \(\int\) Q. (I.F.) dy + c

Where, integrating factor,

I.F. = \(e^{\int p\, dy}\)

Given differential equation is

Equation (1) is of the form

\(\frac{dx}{dy} \, + Px \, =Q\)

Therefore, integrating factor is

General solution is

Putting x=0 and y=0

Therefore, general solution is