Q. 17

# A homogeneous differential equation of the from can be solved by making the substitution.A. (4x + 6y + 5)dy – (3y + 2x + 4)dx = 0B. (xy)dx – (x3 + y3) dy = 0C. (x3 + 2y2)dx + 2xy dy = 0D. y2dx + (x2 – xy – y2)dy = 0

(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 – (x3 + y3)dy = 0

It cannot be homogeneous as the xy which multiplies with dx and x3 + y3 which multiplies with dy are not of same degree.

(C) (x3 + 2y2)dx + 2xy dy = 0

Similarly, it cannot be homogeneous as the x3 + 2y2 which multiplies with dx and 2xy which multiplies with dy are not of same degree.

(D) y2dx + (x2 – xy – y2)dy = 0

Here, putting x = kx and y = ky

= k0.f(x,y)

Therefore, the given differential equations is homogeneous.

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