Answer :

**Given,**

**To Find:** Find the solution when y = 1 and x = 1

**Explanation:**

It can be written as:

…(i)

Let, F(x, y) =

Now, put x = λx and y = λy in F(x, y)

Taking λ as common from both numerator and denominator

We know, If the degree of the λ in function F(x, y) then it is said Homogenous function

So, the given differential equation is Homogenous differential equation.

Put y = vx in equation(i)

Differentiation y w.r.t x , we get

…(iii)

Now, Compare the equation (i) and (iii), we get

Taking x^{2} as common from both numerator and denominator

Integrating both sides, we get

Let v^{2} + 1 = t

On differentiating we get,

2v dv = dt, So

log t = - log x + log C

Putting the value of t , we get

Log|v^{2} + 1| + log|x| = log|c|

log|x(v^{2} + 1)| = log|C|

Now, putting back the value of v = y/x ,

- - - (iv)

When, x = 1 and y = 1

(1)^{2} + 1 = C(1)

C = 2

Put the value of C in equation (iv)

y^{2} + 1 = x(2)

y^{2} = 2x - 1

**Hence,** y^{2} = 2x - 1 is the solution of given differential equation.

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