Answer :

**Given,** A differential equation x^{2} dy + y(x + y)dx = 0

**To Find:** Find the particular solution at y = 1 and x = 1

**Explanation:** **We have given**

x^{2} dy + y(x + y)dx = 0

x^{2} dy = - y(x + y)dx

…(i)

Let F(x, y) =

To check that, given differential equation is homogenous ,

Put x = λx and y = λy in F(x, y)

Then,

Now, Taking λ^{2} common from both numerator and denominator

In F(x, y) If λ^{0}, then F(x, y) is a homogenous function of degree zero.

Let’s put y = vx in equation(i)

On differentiating y , we get

…(ii)

Now, Compare the equation (i) and (ii)

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

Now, Integrating both sides

Solve dv by completing the square method

We know, , then

Now, putting

Now, Put x = 1 and y = 1

1 = 3C

Put the value of C

**Hence,**

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