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 x2 as common from both numerator and denominator
Integrating both sides, we get
Let v2 + 1 = t
On differentiating we get,
2v dv = dt, So
log t = - log x + log C
Putting the value of t , we get
Log|v2 + 1| + log|x| = log|c|
log|x(v2 + 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)
y2 + 1 = x(2)
y2 = 2x - 1
Hence, y2 = 2x - 1 is the solution of given differential equation.
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