Solve the differential equation :

(tan^-1 y – x)dy = (1 + y²)dx.

OR

Find the particular solution of the differential equation

given that y = 1, when x = 0.

Given,

(tan^-1 y – x)dy = (1 + y²)dx

⇒

Clearly, this is a linear differential equation. Comparing with the standard form

A solution of such equation is given by:

x(I.F) = where I.F = integrating factor

We get P(y) = & Q(y) =

Integrating factor I.F is given :

∴ I.F =

We know that:

∴ I.F =

∴ The solution is given as:

⇒ …(1)

Where I =

Let tan^-1y = u

⇒ du =

∴ I =

Using integration by parts:

I =

∴ I =

⇒ I =

∴ the solution is given using equation 1:

⇒

OR

Given,

…(1)

Clearly, the equation is homogeneous(can be observed directly)

We know that for solving a homogeneous we differential equation,

We take y = vx.

As y = vx

Differentiating both sides w.r.t x we get

⇒

∴ equation 1 can be rewritten as-

⇒

∴

⇒

Integrating both sides we get-

⇒

⇒

⇒

⇒

⇒

The above equation gives the general solution. For a particular solution, we need to find the value of C.

As given that at x = 0 ,y = 1

∴

∴ C = 0.

Particular solution at x = 0 and x =1 is given by: