Differential Equations Represented by Dr. Shorouk Ossama

(2) Homogeneous Differential Equation Definition (1): The function of two variables F (x, y) is called homogeneous of degree n if it satisfies that: F (tx, ty) = tn F (x, y) by put x = tx & y = ty   According the definition the function F (x, y) = ax2 + bxy + cy2 is homogeneous of second degree, but function F (x, y) = ax2 + bx + c is non-homogeneous function.

Definition (2): The differential equation M (x, y) dx + N (x, y) dy = 0 is called homogeneous differential equation if M (x, y), N (x, y) are homogeneous of the same degree. Illustrated examples: (i) (x + y) dx + 3x dy = 0 is homogeneous differential equation where, M (x, y) = ( x+ y), N (x, y) = 3x are homogeneous function of the same degree. (M = tx + ty = t(x+y), N = 3x = 3tx)

(ii) (x2 + 5y2) dx + 3xy dy = 0 is homogeneous differential equation where: M (x, y) = (x2 + 5y2) = t2 (x2 + 5y2) , N (x, y) = 3xy = t2(3xy) are homogeneous function of the same degree. (iii) (x2 + y2) dx + sin x dy = 0 is not homogeneous differential equation.

The homogeneous differential equation will be variables separable with substitution, y = ux Conditions For Homogenous are: 1- proves that the equation is Homogenous by put each x = tx and y = ty 2- but each y = ux , it is Separable

Example: Solve the differential equation: Solution The given equation is homogeneous differential equation because, M (x, y) = ( x- y), M = tx – ty = t (x – y) = t M (x, y) and N (x, y) = ( x+ 2y), N (tx, ty) = tx + 2ty = t (x + 2y) = t N (x, y). So M (x, y), N (x, y) are homogeneous of the same degree (first degree).

Let y = ux so dy = u dx + x du Substitute in the differential equation we have: ( x – y) dx + (x + 2y) dy = 0 ( x – ux) dx + (x + 2ux) (u dx + x du) = 0 X (1 – u ) dx + x (1 + 2u) (u dx + x du) = 0

Divided by x (1 – u ) dx + (1 + 2u) (u dx + x du) = 0 (1 – u ) dx + u (1 + 2u) dx + x (1 + 2u) du = 0 (1 – u ) dx + (u + 2u2) dx + x (1 + 2u) du = 0 (1 – u + u + 2u2) dx + x (1 + 2u) du = 0 (1 + 2u2) dx + x (1 + 2u) du = 0

(1 + 2u2) dx + x (1 + 2u) du = 0 Change To Separate variable

Substitute by then the general solution is:

Example: Solve the differential equation: Solution The given equation is homogeneous differential equation because, M (x, y) = y2,M (tx, ty) = t2y2 = t2 M (x, y) and N (x, y) = , N (tx, ty) = t2x2 + txty + t2y2 = t2 N (x, y). So M (x, y), N (x, y) are homogeneous of the same degree (second degree).

Let y = ux so dy = u dx + x du Substitute in the differential equation we have Divided by x2 we have:

Separate variable By integrate the last equation:

Integrate the second term by partial fraction: Where: A, B, C, are constant determined from the equation

When u = -1 then 1 = -c so, C = -1 By comparing the coefficients of the power of u in both sides we have Coefficient of u0: give 1 = A, coefficient of u1: give 1 = 2A + B, B = -1 Substitute in the equation

ln x + ln u – ln (1 + u) + (1 + u)-1 = c

SUMMARY From Pages 21 To 24 Page 25: Exercises

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