For each of the differential equations in question, find the particular solution satisfying the given condition:

(x + y)dy + (x – y) dx = 0; y = 1 when x = 1

(x + y)dy + (x - y)dx = 0

Here, putting x= kx and y = ky

= k^{0}.f(x,y)

Therefore, the given differential equation is homogeneous.

(x + y)dy + (x - y)dx = 0

To solve it we make the substitution.

y = vx

Differentiating eq. with respect to x, we get

Integrating both sides, we get

y = 1 when x = 1

The required solution of the differential equation.

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