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17

A homogeneous differential equation of the from can be solved by making the substitution.

(A) (4x + 6y + 5)dy – (3y + 2x + 4)dx = 0

It cannot be homogeneous as we can see that (4x + 6y + 5) is not homogeneous.

(B) (xy)dx – (x^{3} + y^{3})dy = 0

It cannot be homogeneous as the xy which multiplies with dx and x^{3} + y^{3} which multiplies with dy are not of same degree.

(C) (x^{3} + 2y^{2})dx + 2xy dy = 0

Similarly, it cannot be homogeneous as the x^{3} + 2y^{2} which multiplies with dx and 2xy which multiplies with dy are not of same degree.

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

Here, putting x = kx and y = ky

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

Therefore, the given differential equations is homogeneous.

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